Solution:

Equation (1) is satisfied precisely by the functions $f_1: x \rightarrow -1$ and $f_2: x \rightarrow x+1$. By substitution it is easily confirmed that both functions are solutions.

Now let $f$ be a function that satisfies (1) for all $x, y \in \mathbb{Z}$. Substituting $x=0$ and $y=f(0)$, we obtain, with $z=-f(f(0))$, that $f(z)=-1$. Substituting $y=z$ in (1) leads to $f(x+1)=f(f(x))$ for all $x \in \mathbb{Z}$.

With this we simplify (1) to
$$f(x-f(y)) = f(x+1) - f(y) - 1$$

With $y=x$ in (3) and with (2), we have $f(x+1)-f(x)=f(x-f(x))+1=f(f(x-1-f(x)))+1$.

Since it follows from (3) that $f(x-1-f(x))=f(x)-f(x)-1=-1$, this simplifies to $f(x+1)=f(x)+f(-1)+1=f(x)+c$ with constant $c$. Hence $f$ is linear and satisfies the ansatz $f(x)=c x+b$ with $b=f(0)$.

Substituting this into (2) gives $c x+c+b=c^2 x+c b+b$ for all $x \in \mathbb{Z}$. For $x=0$ and $x=1$ we obtain $c+b=c b+b$ as well as $c^2=c$; from this it follows that $c=0$ or $c=1$. From $c=1$ it follows that $b=1$ and we obtain $f_2$; from $c=0$ it follows that $f$ is constant, and with (1) we obtain $b=-1$, hence $f_1$. This proves everything.