To solve this problem, we are tasked with finding all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that for all integers $a$ and $b$, the following functional equation holds:

$$f^{a^2 + b^2}(a + b) = af(a) + bf(b).$$

We investigate this problem by considering two potential solutions:

1. First Solution: $f(x) = 0$ for all $x \in \mathbb{Z}$.

- If $f(x) = 0$, then substituting into the functional equation gives:
$$f^{a^2 + b^2}(a + b) = 0 = a \cdot 0 + b \cdot 0 = 0.$$
- This identity holds for all integers $a$ and $b$, so $f(x) = 0$ for all $x \in \mathbb{Z}$ is a valid solution.

2. Second Solution: $f(x) = x + 1$ for all $x \in \mathbb{Z}$.

- Substitute $f(x) = x + 1$ into the functional equation:
$$f(a+b) = (a + b) + 1 = a + b + 1.$$
- Compute $af(a) + bf(b)$ with $f(a) = a + 1$ and $f(b) = b + 1$:
$$af(a) + bf(b) = a(a + 1) + b(b + 1) = a^2 + a + b^2 + b.$$
- Check the equation:
$$f^{a^2 + b^2}(a + b) = a^2 + b^2 + a + b = a(a + 1) + b(b + 1).$$
- With this configuration, the equation holds true. Therefore, $f(x) = x + 1$ is another valid solution.

Conclusively, these observations prove that the solutions to the functional equation are:

$$f(x) = 0 \quad \text{for all } x \in \mathbb{Z} \quad \text{and} \quad f(x) = x + 1 \quad \text{for all } x \in \mathbb{Z}.$$

Thus, the solutions are:
$$\boxed{f(x) = 0 \text{ for all } x \in \mathbb{Z} \quad \text{and} \quad f(x) = x + 1 \text{ for all } x \in \mathbb{Z}.}$$