Example 3 Find all pairs of positive integers $(m, n)$ such that
$$n^{5}+n^{4}=7^{m}-1$$
Solving, by rearranging and factoring (1), we get
$$\begin{aligned}
7^{m} & =n^{5}+n^{4}+1=n^{5}+n^{4}+n^{3}-\left(n^{3}-1\right) \\
& =n^{3}\left(n^{2}+n+1\right)-(n-1)\left(n^{2}+n+1\right) \\
& =\left(n^{3}-n+1\right)\left(n^{2}+n+1\right) .
\end{aligned}$$

From (1), we know $n>1$, and when $n=2$, we get $m=2$.
Next, consider the case where $n>2$. We first look at the greatest common divisor of the two expressions on the right side of (2).
$$\begin{aligned}
& \left(n^{3}-n+1, n^{2}+n+1\right)=\left(n^{3}-n+1-\left(n^{2}+n+1\right)(n-1), n^{2}+n+1\right) \\
= & \left(-n+2, n^{2}+n+1\right)=\left(-n+2, n^{2}+n+1+(-n+2)(n+3)\right) \\
= & (-n+2,7) .
\end{aligned}$$

Thus, $\left(n^{3}-n+1, n^{2}+n+1\right) \mid 7$.
Combining this with (2), we know that $n^{3}-n+1$ and $n^{2}+n+1$ are both powers of 7, and when $n \geqslant 3$, they are both greater than 7, which implies $7^{2} \mid\left(n^{3}-n+1, n^{2}+n+1\right)$, leading to a contradiction with the previous result.

In conclusion, the only solution is $(m, n)=(2,2)$.