imc

Clearly, $m=0,n=0$ is one of the solutions. However, we can be quite sure that there are more, so we apply Simon's Favorite Factoring Trick to get the following: $$\begin{array}{rl}{m}^2+mn+n^2& =m^2{n}^2\\ m^2+mn+n^2+mn& =m^2{n}^2+mn\\ (m+n{)}^2& =m^2{n}^2+mn\\ (m+n{)}^2& =mn(mn+1).\end{array}$$ Essentially, this says that the product of two consecutive numbers $mn,mn+1$ must be a perfect square. This is practically impossible except $mn=0$ or $mn+1=0$. $mn=0$ gives $(0,0)$. $mn=-1$ gives $(1,-1),(-1,1)$. Answer: $\boxed{3}.$