16. (57th Belarusian Mathematical Olympiad) Find all positive integers $n, m$, satisfying
$$
n^{5}+n^{4}=7^{m}-1 \text {. }
$$
16. The original equation is equivalent to
$$
\left(n^{3}-n+1\right)\left(n^{2}+n+1\right)=7^{m} \text {. }
$$

Obviously, $n \neq 1$.
When $n=2$, $m=2$.
When $n \geqslant 3$,
$$
\begin{array}{l}
n^{3}-n+1=n\left(n^{2}-1\right)+1>1, \\
n^{2}+n+1>1 .
\end{array}
$$

Let $n^{3}-n+1=7^{a}, n^{2}+n+1=7^{b}$, where $a, b$ are positive integers. Thus, $(n-1)\left(7^{b}-1\right)=7^{a}-1$, which means $\left(7^{b}-1\right) \mid\left(7^{a}-1\right)$.
Let $a=b q+r$, where $q$ is a positive integer, $r$ is a non-negative integer, and $0 \leqslant r < b$.
If $r \neq 0$, then $7^{r} \equiv 1 \pmod{7^{b}-1}$, which is a contradiction.
When $k \geqslant 2$, we have
$$
\left(n^{3}-n+1\right)-\left(n^{2}+n+1\right)^{k} \leqslant\left(n^{3}-n+1\right)-\left(n^{2}+n+1\right)^{2}=-n^{4}-n^{3}-3 n^{2}-3 n<0,
$$

which is a contradiction.
In conclusion, $n=2, m=2$ is the only solution to the original equation.