The most general definition of $\textrm{Ext}^*$ groups for an abelian category $\mathcal{A}$ goes via the derived category $\mathbf{D} (\mathcal{A})$.
Very briefly, the derived category $\mathbf{D} (\mathcal{A})$ is obtained by freely inverting the quasi-isomorphisms in the category of chain complexes $\textbf{Ch} (\mathcal{A})$, and $\textrm{Ext}^n (X, \Sigma^n Y)$ is defined to be the group of morphisms $X \to \Sigma^n Y$ in $\mathbf{D} (\mathcal{A})$, where $X$ is identified (by abuse of notation) with the chain complex concentrated in degree $0$ with value $X$, and $\Sigma^n Y $ is the chain complex concentrated in (homological) degree $n$ with value $Y$.
Note that the general definition makes sense even without injective resolutions, and the relationship with Yoneda extensions is fairly direct: given an exact sequence
$$0 \longrightarrow Y \longrightarrow Z_n \longrightarrow \cdots \longrightarrow Z_1 \longrightarrow X \longrightarrow 0$$
we have a chain complex $Z$ (with $Z_0 = X$, $Z_i$ as given for $1 \le i \le n$ and $Z_i = 0$ otherwise) quasi-isomorphic to $\Sigma^n Y$ and an evident chain complex morphism $X \to Z$, hence an element of $\textrm{Ext}^n (X, Y)$ as desired.
Diagrammatically:
$$\require{AMScd}
\begin{CD}
0 @>>> 0 @>>> 0 @>>> \cdots @>>> 0 @>>> X @>>> 0 \\
@VVV @VVV @VVV & & @VVV @VVV @VVV \\
0 @>>> Z_n @>>> Z_{n-1} @>>> \cdots @>>> Z_1 @>>> X @>>> 0 \\
@AAA @AAA @AAA & & @AAA @AAA @AAA \\
0 @>>> Y @>>> 0 @>>> \cdots @>>> 0 @>>> 0 @>>> 0
\end{CD}
$$
Conversely, given an element of $\textrm{Ext}^n (X, Y)$, we can construct a diagram of the above form and thus obtain a Yoneda extension of $Y$ by $X$, but this requires a bit more theory to justify.
Using the above definitions, it is basically immediate that when we have an exact functor $F : \mathcal{A} \to \mathcal{B}$ between abelian categories, we have an induced functor $\mathbf{D} (\mathcal{A}) \to \mathbf{D} (\mathcal{B})$ and hence natural maps $\textrm{Ext}^n (X, Y) \to \textrm{Ext}^n (F (X), F (Y))$ for all $X$ and $Y$ in $\mathcal{A}$ and all $n$.
However, as mentioned in the comments, even if $F : \mathcal{A} \to \mathcal{B}$ is fully faithful, this map need not be an isomorphism; equivalently, the induced functor $\mathbf{D} (\mathcal{A}) \to \mathbf{D} (\mathcal{B})$ need not be fully faithful.
(For example, take $\mathcal{A}$ to be the category of $\mathbb{F}_p$-modules, take $\mathcal{B}$ to be the category of $\mathbb{Z}$-modules, and take $F$ to be the forgetful functor.)
In general, we must instead consider derived functors, and this is where things start getting really tricky.
It is possible to define right derived functors without assuming enough injectives.
(With that definition, it is a theorem that injective resolutions give the correct answer.)
If we have a right derived functor $\mathbf{R} F : \mathbf{D} (\mathcal{A}) \to \mathbf{D} (\mathcal{B})$ (or even a partial one) we get maps $\textrm{Ext}^n (X, Y) \to \textrm{Ext}^n (\mathbf{R} F (X), \mathbf{R} F (Y))$ and also maps $F (Z) \to \mathbf{R} F (Z)$.
These can be combined to form a zigzag:
$$\textrm{Ext}^n (X, Y) \longrightarrow \textrm{Ext}^n (\mathbf{R} F (X), \mathbf{R} F (Y)) \longleftarrow \textrm{Ext}^n (\mathbf{R} F (X), F (Y)) \longrightarrow \textrm{Ext}^n (F (X), F (Y))$$
In the case where $F (Y) \to \mathbf{R} F (Y)$ is an isomorphism in $\mathbf{D} (\mathcal{B})$ (e.g. if $Y$ admits a resolution by injectives in $\mathcal{A}$ and $F$ sends that to a (not necessarily injective) resolution in $\mathcal{B}$), we get what could be called a canonical map $\textrm{Ext}^n (X, Y) \to \textrm{Ext}^n (F (X), F (Y))$, but I do not know how useful this is.
