Example 4 Find all pairs of positive integers $(m, n)$ such that $m^{6}=n^{n+1}+n-1 .{ }^{[6]}$
(2013, Turkey National Team Selection Exam)
Solve: First, use common methods for solving indeterminate equations (simple modulo, factorization, inequalities, etc.) to eliminate simple cases.
When $n=1$, $m=1$.
Below, assume $n>1$, then $m>1$.
(1) $2 \mid (n+1)$ or $3 \mid (n+1)$.
If $2 \mid (n+1)$, let $n+1=2k$.
Then $\left(m^{3}+n^{k}\right)\left(m^{3}-n^{k}\right)=n-1$.
Note that, $n-1 \neq 0$.
Thus, $n-1=\left(m^{3}+n^{k}\right)\left(m^{3}-n^{k}\right) \geqslant m^{3}+n^{k}>n$.
Contradiction.
Similarly, when $3 \mid (n+1)$, it also leads to a contradiction.
(2) $3 \mid n$.
Then $m^{6} \equiv -1 \pmod{3}$.
Note that, $m^{6}$ is a perfect square.
Thus, $m^{6} \equiv 0$ or $1 \pmod{3}$. Contradiction.
(3) The remaining case is that $n+1$ is a positive integer of the form $6k-1$. This is the most difficult case, consider quadratic residues.
$$
\begin{array}{l}
\text { Then } m^{6}=(n+1-1)^{n+1}+(n+1)-2 \\
\equiv(-1)^{n+1}-2 \equiv -3 \pmod{n+1} \\
\Rightarrow (n+1) \mid \left(m^{6}+3\right).
\end{array}
$$

Since $n+1$ is a positive integer of the form $6k-1$, there is at least one prime factor $p$ of the form $6k-1$ such that
$$
p \mid \left(m^{6}+3\right).
$$

By Theorem 5, this is impossible, contradiction.
In summary, the only positive integer solution to the equation is $m=n=1$.