We provide a combinatorial proof of the given fact. The proof technique uses a combinatorial interpretation for the Gaussian binomial coefficient which you should try and match with yours. Use the Wikipedia page for the $q$-binomial coefficient to connect the inversions and lattice path interpretations. Use this page to connect the lattice path and the tiling interpretation. I won't be using any of these, but I can map my interpretation to the lattice path one easily.
The following interpretation will be used. Let $[n] = \{1,2,\ldots,n\}$ for $n \geq 1$. Given $n,k \geq 1$, let $S^n_k$ be the set of subsets of $[n]$ that have size $k$ (There will be no such subset for $k>n$). Given $S \in S^n_k$, define the weight of $S$ to be $$
w(S) = q^{\sum_{i \in S}\#\{j \notin S : j < i\}}
$$
For example, the weight of $\{2,4\}$ is $q^{1+2}= q^3$. The weight of $\{1,2,3,5\}$ is $q^{0+0+0+1} = q^1$. Then, we have $$
\binom{n}{k}_q = \sum_{S \in S^n_k} w(S)
$$
Here is the appropriate link to the lattice path interpretation. For $S \in S^n_k$, we must have $n \geq k$ (otherwise $S^n_k$ is empty). Consider the lattice $\mathbb Z^2$, and recall the lattice path interpretation : for every lattice path from $(0,0)$ to $(k,n-k)$, the weight of the lattice path is found by calculating the total number of squares that are on the left side of the path, and raising $q$ to that power. We then sum over all lattice paths to get $\binom{n}{k}_q$.
In this case, given $S$, go up if that particular element is in $S$, and go right otherwise. This gives a lattice path. You can easily go backwards. For example, say you're given $\{1,2,3,5\} \in S^{5}_4$. Then, the path is up-up-up-right-up, so you get $(0,0) \to (1,0) \to (2,0) \to (3,0) \to (3,1) \to (4,1)$ There's only one box to the left of this path, so the weight is $1$, as desired.
If you're given $\{2,4\} \in S^{4}_2$, then the path is right-up-right-up, so $(0,0) \to (0,1) \to (1,1) \to (1,2) \to (2,2)$. This will have three blocks to its right, as desired.
For an example the other way, let the path be up-up-right-up-right-up, then the set is $\{1,2,4,6\} \in S^6_4$. You can check that the weights of both equal $3$.
Thus, this bijection is well defined.
We now use the given definition of the $q$-binomial coefficient to write $$
\sum_{i=0}^n (-1)^i q^{\frac{i(i+1)}{2}} \binom{n}{i}_q \binom{n+m-i}{n}_q = \sum_{i=0}^n \sum_{S \in S^n_i} \sum_{S' \in S^{n+m-i}_n} (-1)^i q^{\frac{i(i+1)}2} w(S)w(S')
$$
Now, what we will do is write down a map $f$ that will behave as a signed involution. For $i \geq 1$, let $\mathcal S_i = S^n_i \times S^{n+m-i}_n$. For $i=0$, let $\mathcal S_0 = (\emptyset \times (S^{n+m}_n)) \setminus (\emptyset, [n])$. We have removed the element $(\emptyset, [n])$. If you observe, when $S = \emptyset, S' = [n]$ then ($i=0$, and) the contribution of this term to the sum in the summation above is exactly $q^0 = 1$. This the only term with the contribution $1$, hence we exclude it and focus on the rest.
We will find a map $$
f : \bigcup_{i=0}^n \mathcal S_i \to \bigcup_{i=0}^n \mathcal S_i
$$
Such that :
$f(f(x)) = S$ for all $S \in \bigcup_{i=0}^n \mathcal S_i$.
For $(S,S') \in \bigcup_{i=0}^n \mathcal S_i$, if we let $W(S,S') = q^{\frac{(|S|)(|S|+1)}{2}} w(S)w(S')$, then $W(S,S') = W(f(S,S'))$ for all $(S,S')$.
For $(S,S') \in \bigcup_{i=0}^n \mathcal S_i$, if $f(S,S') = (S_1,S'_1)$ then $|S| - |S_1| = \pm 1$.
Assuming the existence of such a map, $f$ is a bijection but it inverts the sign of the weight because of the last bullet point. Hence, $$
\sum_{i=0}^n \sum_{S \in S^n_i} \sum_{S' \in S^{n+m-i}_n} (-1)^i q^{\frac{i(i+1)}2} w(S)w(S') = 1 + \sum_{(S,S') \in \bigcup_{i=0}^n \mathcal S_i} (-1)^{|S|}W(S,S') \\
= 1 - \sum_{(S,S') \in \bigcup_{i=0}^n \mathcal S_i} (-1)^{|S|}W(S,S') \\
=1
$$
Therefore, all we need to do is construct such a map $f$. While the construction is somewhat unobvious, it is worth keeping in mind that I came up with the idea by observing terms which cancel for small values of $n$ and $m$, followed by the creation of $f$.
Anyhow, here's the description of $f$. Given $(S,S')$, consider the following quantities :
The smallest element of $S$, call it $\min S$. Let $\min \emptyset = +\infty$.
Let $\max S'$ be the largest element of $S'$. Suppose that $r_{S'}$ is the largest number such that all of the numbers $\max S', \max S'-1, \ldots, \max S' - r+1$ are all in $S'$. We call $r$ as the "run size" of $S'$. For example, the run size of $\{1,2,3,5\}$ is $1$ because the number below $5$ is not part of this set. On the other hand, the run size of $\{2,3,6,7\}$ is $2$, the run size of $\{1,3,4,5\}$ is $3$, and so on.
We break the analysis into two cases : either the $r_{S'} < \min S$, or $r_{S'} \geq \min S$.
If $r_{S'} < \min S$, then we describe $f(S,S') = (S_1,S'_1)$. We have $S_1 = S \cup \{r_{S'}\}$, and $S'_1 = (S_1 \setminus \{\max S_1\}) \cup \{\max S_1 - r_{S'}\}$.
Suppose $r_{S'} \geq \min S$,then we describe $f(S,S') = (S_1,S'_1)$. In this case, $S_1 = S \setminus \{\min S\}$ and $S'_1 = (S' \setminus \{\max S' - \min S + 1\}) \cup \{\max S' + 1\}$.
Let me give some examples of this bijection to make it utterly clear what is going on.
Suppose that $S = \emptyset$ and $S' = \{2,4,5\}$. Then $r_{S'} = 2$ and $\min S = +\infty$. We are as in the first case, so $S_1 = \emptyset \cup \{2\} = \{2\}$, and $S'_1 = \{2,4,5\} \setminus \{5\} \cup \{3\} = \{2,3,4\}$. So, $f(\emptyset, \{2,4,5\}) = (\{2\},\{2,3,4\})$.
To show the involutory property, let's just go the other way. Suppose that $S = \{2\}$ and $S' = \{2,3,4\}$. Then, $r_{S'} = 3$ and $\min S = 2$. We are in the second case, so $S_1$ comes by removing $\min S$ i.e. $S_1 = \emptyset$ and $S'_1$ comes from removing $4-2+1 = 3$ and adding $4+1 = 5$, so that we get back $S'_1 = \{2,4,5\}$.
A more complicated example : Suppose that $S = \{3,7\}$ and $S' = \{2,4,5,6,8,9,10\}$. Then, $r_{S'} = 3$ and $\min S = 3$. We are in the second case, so $f(S,S') = (S_1,S'_1)$ where $S_1$ comes from removing the minimum i.e. $S_1 = \{7\}$ and $S'_1$ comes from removing $10-3+1$ i.e. $8$ and replacing it by $10+1$ i.e. $11$ so that $S'_1 = \{2,4,5,6,9,10,11\}$.
Let's now prove the properties of $f$ that we asserted. Given a pair $(S,S')$, suppose that it fits the first case i.e. $r_{S'} < \min S$. Then, following the procedure to get $S_1,S'_1 = f(S,S')$, by definition we have $S_1 = S \cup \{r_{S'}\}$ and $S'_1 = S \setminus \{\max S_1\} \cup \{\max S_1 - r_{S'}\}$. Following this, observe that $\min S_1 = r_{S'}$. We have $\max S'_1 = \max S' - 1$, and because the elements in the "run" of $S'$ were kept on in $S_1$, we actually added to that run by adding the element $\max S_1 - r_{S'}$. In other words, one can check that $r_{S'_1} \geq r_{S'} = \min S_1$. Therefore, we are in the second case, and one easily checks by going back that $f(f(S,S')) = (S,S')$, whenever $(S,S')$ are as in the first case. If they are in the second case, a similar analysis holds.
The third bullet point is obvious, since while forming $f(S,S') = (S_1,S'_1)$, $S_1$ is formed by either removing an element in $S$ or inserting an element into $S$ that is not in $S$.
The second bullet point is proved by cases. Let us illustrate this by an example.
If $(S,S') = (\{2,3\}, \{4,5,7\})$ then $f(S,S') = (\{1,2,3\},\{4,5,6\})$. We have $W(S,S') = q^{3 + 2+10} = q^{15}$ and $W(f(S,S')) = q^{6+0+9} = q^{15}$ as well. Observe that the weight of the $S$ component went down by $2 = |S|- r_{S'}+1$, and the weight of the $S'$ component went down by $1= r_{S'}$. The total decrease is by $|S|+1$. However, this is compensated by the fact that the power of $q$ related to $|S|$ went up by exactly that amount.
Suppose that $(S,S')$ are as in the first case i.e. $r_{S'} < \min S$. Then, if $f(S,S') = (S_1,S'_1)$, we see that $$w(S_1) = w(S \cup \{r_{S'}\}) = w(S)q^{(r_{S'}-1)-(|S_1|-1)} = w(S)q^{r_{S'}-|S|-1}$$. This is because when we add $r_{S'}$, then $r_{S'} = \min S_1$ by definition, so this adds $r_{S'}-1$ to the power in $q$. On the other hand, every other element in $S'$ is bigger than $r_{S'}$ so when $r_{S'}$ is added, the weight corresponding to that element decreases from the weight corresponding to $S$ by $1$ ,amd there are $|S|$ such elements.
On the other hand, $$
w(S'_1) = w(S' \setminus \{\max S'\}) \cup \{\max S' - r_{S'}\}
$$
When we replace $\max S'$ by $\max S_1 -r_{S'}$, then this costs one power of $q$. For every element between these as well, one power of $q$ is removed. Thus, in total, a total exponent of $r_{S'}$ is lost. All in all, we get $w(S'_1) = w(S')q^{-r_S}$.
Therefore, $$
w(S'_1)w(S_1) = w(S)w(S')q^{-(|S|+1)}\\
\implies W((S,S')) = w(S)w(S')q^{|S|(|S|+1)/2} = w(S)w(S')q^{(|S|+1)(|S|+2)/2}q^{-(|S|+1)} = w(S_1)w(S'_1)q^{(|S'|)(|S'|+1)/2} = W(f(S,S'))
$$
as desired. Something similar applies to the second case.
Note that this proof may have been somewhat surprising, especially for those who are used to more power-series based proofs. Another important point is that this proof works even when $m<n$, if we assume that $\binom{n}{k}_q = 0$ for $k>n$. The same proof works, except that you only need to consider $\mathcal S_i$ for $i=0,1,\ldots,\min\{m,n\}$. That is, we get $$
\sum_{i=0}^{\min\{m,n\}} (-1)^i q^{i(i+1)/2}\binom{n}{i}_q \binom{m+n-i}{n}_q = 1
$$
