What is the set of functions $f:\mathbb{Z} \rightarrow \mathbb{Z}$ such that $(f(x+y))^2 = f(x^2) + f(y^2)$ for all $x, y, \in \mathbb{Z}$.
I've computed that $f(z) = 0$ and $f(z) = 1$ are the only such functions, for all $z \in \mathbb{Z}$, since
$$ f(0)(f(0) - 2) = 0 $$ and $$ f(x^2) = \frac{1}{2}(f(0))^2 $$
Is there anything else I'm missing here?
Not quite right.
Put $x = y =0$ we have $f(0)^2 = 2f(0)$, so $f(0) =0$ or $f(0) =2$. Put $x =-y$, then $f(0)^2 = 2f(x^2)$ for all $x\in \Bbb Z$.
When $f(0) =0$, $f(x^2 ) = f(0) =0$. Put $y=0$ we have $f(x)^2 = f(x^2) + f(0) = f(x^2) $, hence $f(x)^2 = 0$ then $f(x) = 0$.
[Critical Part]
When $f(0) = 2$, similarly we have $f(x^2) =2$ and $f(x)^2 = f(x^2) + f(0) = 2+f(x^2)$, then $f(x)^2 = 4$, $f(x) = \color{red}{\pm 2}$. So conclusively we only require that $f(x) = 2$ for perfect squared $x$, i.e. there is a $z\in \Bbb Z$ s.t. $x = z^2$. On the other hand for every $f \colon \mathbb Z \to \mathbb Z$ that $$ f(x) =\begin{cases} 2, & x = z^2, z\in \mathbb Z, \\ \pm 2, & \text { else }, \end{cases} $$ the equation $$ f(x+y)^2 = f(x^2) + f(y^2) $$ becomes $$ (\pm 2)^2 = 2 + 2, $$ which is an identity.
Conclusion: the set of such functions is $$ \left\{ 0\right\} \bigcup \left\{f \colon \mathbb Z \to \mathbb Z\colon f(m^2) = 2 [m \in \mathbb Z]; |f(x)|=2, x \in \mathbb Z \right\}. $$