Example 36 (2005 Slovenia National Team Selection Exam) Find all pairs of positive integers $(m, n)$ such that $m^{2}-4 n$ and $n^{2}-4 m$ are both perfect squares.
Solution: Clearly, $m^{2}-4 n<m^{2}$.
If $m^{2}-4 n=(m-1)^{2}$, then $2 m-1=4 n$, which is a contradiction.
Therefore, $m^{2}-4 n \leqslant(m-2)^{2}$, which implies $4 m \leqslant 4 n+4$, or $m \leqslant n+1$.
Similarly, $n \leqslant m+1$.
Hence, $n-1 \leqslant m \leqslant n+1$.
(1) If $m=n-1$, then
$$
\begin{array}{l}
n^{2}-4 m=n^{2}-4(n-1)=(n-2)^{2}, \\
m^{2}-4 n=m^{2}-4(m+1)=(m-2)^{2}-8=t^{2}\left(t \in \mathbf{N}_{+}\right) .
\end{array}
$$

Thus, $(m-2+t)(m-2-t)=8$.
Hence $\left\{\begin{array}{l}m-2+t=4, \\ m-2-t=2,\end{array}\right.$ solving gives $m-2=3$.
Therefore, $m=5, n=6$, which satisfies the requirement.
(2) If $m=n$, then
$$
m^{2}-4 n=n^{2}-4 m=m^{2}-4 m=(m-2)^{2}-4=t^{2}(t \in \mathbf{N}) \text {. }
$$

Thus $(m-2+t)(m-2-t)=4$, solving gives $m-2=2$.
Therefore, $m=n=4$, which satisfies the requirement.
(3) If $m=n+1$, then
$$
\begin{array}{l}
m^{2}-4 n=(n+1)^{2}-4 n=(n-1)^{2}, \\
n^{2}-4 m=n^{2}-4(n+1)=(n-2)^{2}-8=t^{2}\left(t \in \mathbf{N}_{+}\right),
\end{array}
$$

Thus, $(n-2+t)(n-2-t)=8$,
Solving gives $n-2=3$.
Therefore, $n=5, m=6$, which satisfies the requirement.
In summary, the required pairs of positive integers $(m, n)=(4,4),(5,6),(6,5)$.