Example 1 (Problem from the 35th IMO) Find all ordered pairs of positive integers $(m, n)$ such that $\frac{n^{3}+1}{m n-1}$ is an integer.

Find all ordered pairs of positive integers $(m, n)$ such that $\frac{n^{3}+1}{m n-1}$ is an integer.
Let $n^{3}+1$ be a multiple of $m n-1$, then $\left(n^{3}+1\right)+m n-1=n\left(n^{2}+m\right)$ is divisible by $n m-1$. Since $n$ and $n m-1$ are coprime, $n^{2}+m$ is divisible by $n m-1$. By symmetry, assume $m \geqslant n$.

If $n=1$, then $\frac{n^{3}+1}{m-1}=\frac{2}{m-1}$ is an integer, so $(m-1) \mid 2$, thus $m=2$ or 3.

If $n=m$, then $\left(n^{2}-1\right) \mid\left(n^{2}+m\right)$ implies $(n-1) \mid 1$, hence $n=m=2$.

If $m>n \geqslant 2$, if $n^{2}+m \geqslant 2(n m-1)$, then $n^{2}+n \geqslant m(2 n-1) \geqslant(n+1)(2 n-1)$, which means $3 \geqslant n^{2}+n$ and $n \geqslant 2$, which is impossible. Therefore, only $n^{2}+m=n m-1$, $(n+1)(n-1)+2=m(n-1)$, indicating $(n-1) \mid 2$. Hence $n=2$ or 3, thus $m=5$ or 5.

In summary, all ordered pairs of positive integers $(m, n)$ are 9: $(2,1),(3,1),(1,2),(1,3),(2,2)$, $(5,2),(5,3),(2,5),(3,5)$.