Example 3 Find all positive integers $m, n$ such that $m^{2}-n$ divides $m+n^{2}$, and $n^{2}-m$ divides $n+m^{2}$.
(2002 Asia Pacific Mathematical Competition)
Analysis: By symmetry, we may assume $n \geqslant m$ and estimate the rough range of $n$.
Solution: Without loss of generality, assume $n \geqslant m$, then $(m+1)^{2}-m=(m+1)+m^{2}$. Clearly, $n^{2}-m>n+m^{2}$ holds for $n>m+1$. Therefore, when $n>m+1$, there are no integers $m, n$ that satisfy the condition. We only need to consider $n=m$ and $n=m+1$.

Assume $n=m$. Then $n^{2}-n$ divides $n^{2}+n$. If $n>3$, then $n^{2}>3 n \Rightarrow 2\left(n^{2}-n\right)>n^{2}+n$, clearly $n^{2}-n$ does not divide $n^{2}+n$ for $n>3$. Therefore, $n=2$ and $n=3$ are the only possibilities, and it is easy to verify that they work.
Assume $n=m+1$. We seek $m$ such that $m^{2}-m-1$ divides $m^{2}+3 m+1$. If $m \geqslant 6$, then
$$
m(m-5)>3 \Rightarrow 2\left(m^{2}-m-1\right)>m^{2}+3 m+1 \text{. Also, } m^{2}-m-1<m^{2}+3 m+1 \text{. }
$$

So $m^{2}-m-1$ does not divide $m^{2}+3 m+1$. For $m=1,2,3,4,5$, we check each one and find that $m=1$ or $m=2$. In summary, the solutions are $(m, n)=(2,2),(3,3),(1,2),(2,1),(3,2),(2,3)$.