Suppose that $\tau_1$ and $\tau_2$ are two topologies on a set $X$ with the property that $K\subset X$ is compact with respect to $\tau_1$ if and only if $K$ is compact with respect to $\tau_2$. Then is this enough information to determine whether $\tau_1 = \tau_2$?
If not (which is suspect to be the case) is there a nice counterexample, and what is the minimum amount of extra information required for $\tau_1 = \tau_2$?
I feel like there would be an issue regarding 'points at infinity.'
It is not enough information. A simple (and perhaps frightening) counterexample is as follows: if $X$ is an uncountable set, then in both the discrete and co-countable topologies the compact sets are exactly the finite sets. It's frightening because these two topologies share virtually no properties: one is metrizable (hence perfectly normal), the other isn't even Hausdorff; one is Lindelöf, and the other has maximal Lindelöf number.
I'm not certain what additional information would be required to conclude that two topologies are the same. (I'll think about this some more.)
Addition 1
Hausdorffness is not sufficient. The discrete space on $\mathbb{N}$ and the Arens–Fort space are both Hausdorff, and the compact subsets are exactly the finite subsets, but these spaces are not homeomorphic. (Since they are both countable, we can transfer the topology of one onto the underlying set of the other, and the compact sets will be the same.)
Actually, this example shows that perfect normality, Lindelöfness, and paracompactness are not sufficient.
Consider a finite set $X$,then every subset $A$ of $X$ is compact in the trivial topology $\{\emptyset,X\}$, but every subset $A$ of $X$ is also compact in the discrete topology on X (the topology where all subsets of X are considered open). If $|X|>1$, then these topologies are obviously not equal.
I will return to your second question if it happens that I have anything useful to say about it.
As others pointed out, there are examples of spaces where compact subsets are the same but the spaces aren't homeomorphic. For example anticompact spaces, spaces for which every compact subset is finite, provide some examples (finite spaces, discrete spaces, spaces with cocountable topology, Arens-Fort space, see here for more examples on pi-base). Another class of spaces which give some examples are Noetherian spaces, spaces for which every subspace is compact (finite spaces, indiscrete spaces, spaces with cofinite topology, see here for more examples on pi-base).
Definition. We call $X$ a KC space if every compact set is closed, we call $X$ a compactly generated space (or $k$-space) if $U\subseteq X$ is open iff $U\cap K\subseteq K$ is open in $K$ for all compact $K\subseteq X$.
Compactly generated KC spaces include sequential Hausdorff spaces, e.g. metric spaces. But they also include more exotic topologies like the Alexandroff one point compactification of any metrizable space, which are $T_1$ but generally not $T_2$ (here I show any such space is a KC space, and that they are countably generated follows from compactness).
Let $\mathcal{K}_X, \mathcal{C}_X, \mathcal{KC}_X, \mathcal{F}_X$ denote all compact, all closed, all compact closed, and all finite subsets of $(X, \tau)$. Necessarily $\mathcal{F}_X\subseteq \mathcal{K}_X$ and $\mathcal{KC}_X=\mathcal{K}_X\cap\mathcal{C}_X$, while all the other relations don't hold in general. \begin{align*} \mathcal{F}_X\subseteq \mathcal{C}_X\text{ is equivalent to }T_1,&&\mathcal{K}_X\subseteq \mathcal{C}_X\text{ is equivalent to KC},\\ \mathcal{C}_X\subseteq \mathcal{K}_X\text{ is equivalent to compact,}&& \mathcal{C}_X\subseteq \mathcal{F}_X\text{ is equivalent to finite,}\\ \mathcal{K}_X = \mathcal{F}_X\text{ is equivalent to anticompact,}&&\mathcal{K}_X = 2^X\text{ is equivalent to Noetherian,} \\ \mathcal{C}_X = 2^X\text{ is equivalent to discrete,}&& \mathcal{F}_X = 2^X\text{ is equivalent to finite.}\end{align*}
We know $\emptyset\in \mathcal{F}_X, \mathcal{K}_X, \mathcal{C}_X, \mathcal{KC}_X$ and $X\in\mathcal{C}_X$, and each of the classes $\mathcal{F}_X, \mathcal{K}_X, \mathcal{C}_X, \mathcal{KC}_X$ is closed under finite unions, $\mathcal{F}_X, \mathcal{C}_X, \mathcal{KC}_X$ are closed under arbitrary intersections, while $\mathcal{K}_X$ need not be closed under finite intersections. A sufficient condition for $\mathcal{K}_X$ to be closed under arbitrary intersections is that $X$ is a KC space.
Proposition 1. $X$ is compactly generated KC space iff $\mathcal{C}_X = \{A\subseteq X : \forall_{K\in \mathcal{K}_X} A\cap K\in \mathcal{K}_X\}$ and $\mathcal{K}_X$ is closed under finite intersections
Proof: First note $\mathcal{C}_X\subseteq\{A\subseteq X : \forall_{K\in \mathcal{K}_X} A\cap K\in \mathcal{K}_X\}$ is always true, since if $A$ is closed and $K$ is compact, then $A\cap K$ is closed in $K$, so compact. If $A\cap K$ is compact for all compact $K$, then if $X$ is a KC space, we can say that $A\cap K$ is closed in $X$, and so in $K$. If $X$ is compactly generated, this implies that $A$ is closed in $X$. Now conversely suppose that $\mathcal{C}_X = \{A\subseteq X : \forall_{K\in \mathcal{K}_X} A\cap K\in \mathcal{K}_X\}$. If $A$ is compact, then $A\cap K$ is compact for any compact $K$, so that $A$ is closed. This implies $X$ is a KC space. If $A\cap K$ is closed in $K$ for any compact $K$, then $A\cap K$ is compact for any compact $K$, so that $A$ is closed. So $X$ is compactly generated. $\square$
This proposition shows that if $f:X\to Y$ is a bijection between compactly generated KC spaces, such that $f(\mathcal{K}_X) = \mathcal{K}_Y$, then $f$ is a homeomorphism.
Proposition 2. $X$ is generated by compact closed subsets iff $$\mathcal{C}_X = \{A\subseteq X : \forall_{K\in \mathcal{KC}_X} A\cap K\in \mathcal{KC}_X\}$$
Proof: $\mathcal{C}_X\subseteq\{A\subseteq X : \forall_{K\in \mathcal{KC}_X} A\cap K\in \mathcal{KC}_X\}$ is always true, just like in proposition 1. If $A\cap K$ is compact closed for every compact closed $K$, then $A\cap K$ is closed in every compact closed $K$, so $A$ is closed in $X$ when $X$ is generated by compact closed sets. Conversely, suppose that $A\cap K$ is closed in compact closed $K$. Then $A\cap K$ is compact as a closed subset of compact space, and it's closed in $X$ since $K$ is closed in $X$. So $A\cap K\in\mathcal{KC}_X$ for $K\in\mathcal{K}_X$ and so $A$ is closed. $\square$
Let's investigate conditions under which $\mathcal{K}\subseteq 2^X$ will be a family of compact sets. Necessarily $\mathcal{F}_X\subseteq\mathcal{K}$ and $\mathcal{K}$ is closed under finite unions. Let's try to give topology to $X$ according to the formula $\mathcal{C}_X = \{A\subseteq X : \forall_{K\in \mathcal{K}} A\cap K\in \mathcal{K}\}$. The first problem is that $\mathcal{C}_X$ needs to be closed under arbitrary intersections in order to be a family of closed sets. For this we can assume that $\mathcal{K}$ is closed under arbitrary intersections. Then $X$ defines a $T_1$ topological space. In this topology, there is no guarantee that $\mathcal{K}$ consists of compact sets. This can be fixed by demanding that subfamilies $\mathcal{K}'\subseteq \mathcal{K}$ with finite intersection property have non-empty intersection. This will guarantee that $\mathcal{K}\subseteq \mathcal{K}_X$. Then $X$ will be generated by compact closed sets, since if $A\cap K$ is closed in $K$ for all compact closed $K$, then $A\cap K$ will be closed in $K$ for all $K\in\mathcal{K}$, which is seen to be equivalent to $A\cap K\in \mathcal{K}$ for all $K\in\mathcal{K}$, hence $A\in\mathcal{C}_X$.
We see that perhaps this construction is more suited for $\mathcal{K}$ to be a family of compact closed sets. In this case by not demanding $\mathcal{F}_X\subseteq \mathcal{K}$ we obtain a slightly more general construction.
Suppose you additionally assume that for any $K\in\mathcal{K}$, the topology given by closed sets $\mathcal{C}_K = \{A\in\mathcal{K}: A\subseteq K\}$ is a KC space. This topology on $K$ coincides with subspace topology $K\subseteq X$. If $A\subseteq X$ is compact, then $A\cap K$ is a closed subset of $A$, so compact subset of $K$, so closed in $K$, and hence $A\cap K\in\mathcal{K}$. It follows that $A\in\mathcal{C}_X$ so that $X$ is a KC space.
Note that I haven't claimed that $\mathcal{K} = \mathcal{K}_X$ under these conditions, and it is false since $\mathcal{K} = \{K\subseteq \mathbb{Q}^\ast : K\setminus \{\infty\}\text{ is compact}\}$ where $\mathbb{Q}^\ast$ is the Alexandroff one point compactification of $\mathbb{Q}$ satisfies all the properties but $\mathbb{Q}^\ast\notin \mathcal{K}$ even though $\mathbb{Q}^\ast$ is compact.
Here are another couple of examples.
The first example shows that compact sets do not determine sequential convergence. Define $X := \mathbb{N}\times\mathbb{N}\cup\{(0,0)\}$ where $\mathbb{N} = \{1, 2, \ldots \}$. Let $\tau_{1}$ be the cofinite topology on $X$. We also define a neighbourhood system $\mathcal{N}$ on $X$ as follows. Let $N\subseteq X$.
If $(m,n)\in \mathbb{N}\times\mathbb{N}$ then $N$ is a $\mathcal{N}$-neighbourhood of $(m,n)$ if $(m,n)\in N$ and $N$ is cofinite.
The set $N$ is a $\mathcal{N}$-neighbourhood of $(0,0)$ if $(0,0)\in N$ and the set \begin{equation} \{m\in\mathbb{N} : \{n\in\mathbb{N} : (m,n)\not\in N \} \text{ is finite} \} \end{equation} is cofinite.
It is straightforward to check that this is indeed a neighbourhood system on $X$. Consequently, this neighbourhood system $\mathcal{N}$ defines a topology $\tau_{2}$ on $X$. Note that the neighbourhoods for both topologies coincide at all points apart from $(0,0)$. Hence any subset of $X$ not containing $(0,0)$ has the same induced topology from both the $\tau_{1}$ and $\tau_{2}$ topologies. Note that both topologies are $T_{1}$, neither are Hausdorff and the $\tau_{1}$ topology is coarser than the $\tau_{2}$ topology.
We first show that every subset of $X$ is both $\tau_{1}$-compact and $\tau_{2}$-compact. That every subset of $X$ is $\tau_{1}$-compact follows from the same reasoning as in the answers here. Let $A\subseteq X$. If $(0,0)\not\in A$ then the $\tau_{1}$ and $\tau_{2}$ topologies coincide on $A$ and then $A$ is $\tau_{2}$-compact. So suppose $(0,0)\in A$. Let $\mathscr{U}$ be an open cover of $A$. There is $U\in \mathscr{U}$ with $(0,0)\in U$. Then $\mathscr{U}\setminus\{U\}$ is an open cover for $A\setminus (0,0)$. Since the $\tau_{1}$ and $\tau_{2}$ topologies coincide on $A\setminus (0,0)$, there is a finite set $\mathscr{F}\subseteq \mathscr{U}\setminus\{U\}$ which covers $A\setminus (0,0)$. Then $\mathscr{F}\cup \{U\}$ is a finite subset of $\mathscr{U}$ that covers $A$. Hence $A$ is $\tau_{2}$-compact.
Let $(a_{n})_{n\in\mathbb{N}}$ be an enumeration of $\mathbb{N}\times\mathbb{N}$. We claim that $(a_{n})_{n\in\mathbb{N}}$ converges to $(0,0)$ in the $\tau_{1}$ topology but not in the $\tau_{2}$ topology. From the characterisation of sequences in the cofinite topology discussed here, we see that $(a_{n})_{n\in\mathbb{N}}$ converges to $(0,0)$ in the $\tau_{1}$ topology. On the other hand, using the same reasoning as in this answer we have that $(0,0)$ cannot be the $\tau_{2}$-limit of a sequence in $\mathbb{N}\times\mathbb{N}$. Hence the sequence $(a_{n})_{n\in\mathbb{N}}$ does not converge to $(0,0)$ in the $\tau_{2}$ topology.
It is worth noting that the topologies in the above example were not Hausdorff. If both topologies are Hausdorff and have the same compact sets, then a sequence converges to some limit in one topology if and only if the sequence converges to the same limit in the other topology. See here for details.
The second example is from functional analysis. Let $X$ be an infinite-dimensional Banach space. Let $X^{*}$ denote its topological dual. The weak$^{*}$ topology on $X^{*}$, denoted by ${\rm w}^{*}$, is the coarsest topology which makes the map $Jx\colon X^{*}\to\mathbb{K}$ defined by $(Jx)(f) := f(x)$ continuous for every $x\in X$. The bounded weak$^{*}$ topology on $X^{*}$, denoted by ${\rm bw}^{*}$, is the finest topology which coincides with the weak$^{*}$ topology on each norm bounded subset of $X^{*}$. Note that both topologies are Hausdorff and the weak$^{*}$ topology is coarser than the bounded weak$^{*}$ topology.
Regarding the compact sets in each topology, we have the following result.
Let $B\subseteq X^{*}$. Then the following are equivalent.
${\rm (i)}$ $B$ is norm bounded and weak$^{*}$ closed.
${\rm (ii)}$ $B$ is compact in $(X^{*}, {\rm w}^{*})$.
${\rm (iii)}$ $B$ is compact in $(X^{*}, {\rm bw}^{*})$.
Proof. The implication ${\rm (i)} \Rightarrow {\rm (ii)}$ follows from the Banach-Alaoglu theorem [Corollary 2.6.19 in [1]]. The implication ${\rm (ii)} \Rightarrow {\rm (i)}$ follows from recalling that compact subsets of a Hausdorff topological vector space are closed and bounded, that the weak$^{*}$ topology is Hausdorff and that bounded subsets of $(X^{*}, {\rm w}^{*})$ are norm bounded by the uniform boundedness theorem and the completeness of $X$ [Corollary 2.6.9 in [1]]. Since the ${\rm w}^{*}$ topology is coarser than the ${\rm bw}^{*}$ topology we automatically have $({\rm iii}) \Rightarrow ({\rm ii})$. The implication $({\rm i}) \text{ and } ({\rm ii}) \Rightarrow ({\rm iii})$ follows from recalling that the ${\rm w}^{*}$ and ${\rm bw}^{*}$ topologies coincide on norm bounded sets. This completes the proof.
However, the weak$^{*}$ and bounded weak$^{*}$ topologies on $X^{*}$ are distinct whenever $X$ is infinite-dimensional. See Corollary 2.7.7 in [1].
[1] Megginson, R. E., An introduction to Banach space theory, vol. 183 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1998.
