I think I might be able to provide a proof for a slightly modified version of the hypotheses, which would still be enough for the theorem that I am trying to prove.

Let $F_1:=\{(y,z) \to f_1(y,z)\}$ and $F_2:= \{(y,z) \to (f_y(y),f_z(z))\}$ where $f_1$ and $f_2$ are non-injective. This relaxes the hypothesis that $f_y$ and $f_z$ must also be separately non-injective. We then have
$$
\max_{F_1} \big\{ \ I(X; f_1(Y,Z)) \ \big\} = \max_{F_2} \big \{ \ I(X; f_2(Y,Z)) \ \big \}
$$

**Proof**. Call $f_1^* = \arg \max_{F_1} \big\{ \ I(X; f_1(Y,Z)) \ \big\}$ and $f_2^* = (f_y^*, f_z^*)= \arg \max_{F_2} \big\{ \ I(X; f_2(Y,Z)) \ \big\}$. We have
$$
I(X; Y, Z) \stackrel{(a)}{\ge} I(X; f_1^*(Y,Z)) \stackrel{(b)}{\ge} I(X; f_2^*(Y,Z)) = I(X; f_y^*(Y), f_z^*(Y)) \stackrel{(c)}{\ge} I(X; Y, f_z^*) \stackrel{(d)}{=} I(X;Y)
$$
where (a) follows from the data processing inequality, (b) from the fact that $F_2 \subseteq F_1$, (c) from the definition of $f_2^*$ and from the fact that $(y, f_z^*(z)) \in F_2$, and (d) from the fact that $f_z^*(Z)$ is independent of $X$ and $Y$. Because $I(X;Y,Z) = I(X;Y)$ we obtain $I(X; f_1^*(Y,Z)) = I(X; f_2^*(Y,Z))$. Is that correct?