Suppose I have a function such that $F(\theta x+(1-\theta)y,z)=\theta F(x,z)+(1-\theta)F(y,z)$ for $\theta\in(0,1)$. I want to show that $F(x,z)$ is concave in the first argument when taking a supremum, that is $G(x)=\sup_z F(x,z)\text{ is concave.}$
Let $\theta\in(0,1)$. Then $F(\theta x+(1-\theta)y,z)=\theta F(x,z)+(1-\theta)F(y,z)$ and so $$G(\theta x+(1-\theta)y)=\sup_z F(\theta x+(1-\theta)y,z)\geq\theta F(x,z)+(1-\theta)F(y,z).$$
What I don't know is also how to do is bring the $\sup$ onto the RHS, ie. to get $$G(\theta x+(1-\theta)y)\geq\theta G(x)+(1-\theta)G(y).$$
There is a result that says if $F(x,z)$ is concave in both ($x,z)$, then $\sup_z F(x,z)$ is also concave, but I don't have this luxury of concavity in the second argument. I am also not sure if the result is correct, but was hoping to verify it.
The supremum is actually convex because supremum of a family of convex functions is convex. Your case is almost equivalent to the full generality of this fact and therefore there is no reason to believe that $\sup_zF(x,z)$ is affine.