13. If $n$ is a prime, then when $n \mid m$, it is obvious that $n \nmid\left(m^{n-1}+1\right)$; if $n \nmid m$, then by Fermat's Little Theorem, we know $n \mid\left(m^{n-1}-1\right)$; if $n \mid\left(m^{n-1}+1\right)$, then it requires $n \mid 2$, which contradicts $n$ being an odd number greater than 1.

If $n$ is a composite number, and there exists $m \in \mathbf{N}^{*}$, such that $n \mid\left(m^{n-1}+1\right)$. We set $2^{t} \|(n-1)$, then
$$\left(m^{k}\right)^{2^{t}} \equiv-1(\bmod n)$$

where $k=\frac{n-1}{2^{t}} \in \mathbf{N}^{*}$. For any prime factor $p$ of $n$, by (1) we know $\left(m^{k}\right)^{2^{t}} \equiv-1(\bmod p)$, and by Fermat's Little Theorem, we know $\left(m^{k}\right)^{p-1} \equiv 1(\bmod p)$, so the power of 2 in the prime factorization of $\delta_{p}\left(m^{k}\right)$ is $\geqslant t+1$, and $2^{t+1} \mid(p-1)$. Since $p$ is any prime factor of $n$, this requires $n \equiv 1\left(\bmod 2^{t+1}\right)$, which contradicts the definition of $t$.

Therefore, the proposition holds.