Describe all ring homomorphisms from $\mathbb Z\times\mathbb Z$ to $\mathbb Z\times\mathbb Z$

You know that $(1,0) \cdot (0,1)=(0,0)$ then $f(1,0) \cdot f(0,1)=f(0,0)=0$. This implies that $f(1,0)=(a,0)$ or $(0,a)$, for some $a\in\mathbb{Z}$ and similarly for $f(0,1)=(b,0)$ or $(0,b)$. Since $f(1,1)=(1,1)$ you get that $a=b=1$ and that only two of the four options are valid. In this way you get exactly your two functions.

I have a manual that goes through the solutions to this problem. Although Quimey is on track, there is actually 9 possibilities, and they all describe a ring homomorphism.

Think of it this way. Let $f\colon \mathbb Z\times\mathbb Z \to \mathbb Z\times\mathbb Z$ be the function. Then suppose that $f(1,0) = (m,n)$. Well, $f(1,0) = f( (1,0)(1,0) ) = f(1,0)f(1,0) = (m,n)(m,n) = (m^2,n^2)$. So when is it true in $\mathbb Z$ for $m^2 = m$ and $n^2 = n$? only when $m$ and $n$ are $1$ or $0$.

This means that $f(1,0) = (1,0) , (0,1) , (1,1) , (0,0)$

Notice though, that they have to be in certain combinations with each other because of the reason Quimey stated.