Solution:

The answer is 4. The given equation is equivalent to:
$$\left(x^{2}+y^{2}-1\right)\left(x^{2}+y^{2}-7\right)=0,$$
whose solutions are given by the pairs $(x, y)$ of integers that satisfy $x^{2}+y^{2}=1$ or $x^{2}+y^{2}=7$. The latter equation, however, is impossible, since an integer that is a sum of two squares cannot have remainder 3 upon division by 4. It follows that all and only the solutions are $(\pm 1,0)$ and $(0, \pm 1)$.

(Alfieri)