Set $\alpha = v_p(n-1)$ and write $n-1 = r \cdot p^{\alpha}$. Let $q$ be an arbitrary prime divisor of $n$, and set $d = \text{ord}_q(m)$. Since
$$m^{p(n-1)} = m^{r \cdot p^{\alpha+1}} \equiv 1 \pmod{q},$$
it follows that $d \mid r \cdot p^{\alpha+1}$. If $v_p(d) \neq \alpha + 1$, then clearly $d \mid n-1$, and therefore $q$ is a common divisor of $n$ and $m^{n-1} - 1$. Otherwise, suppose that $v_p(d) = \alpha + 1$ for all prime divisors of $n$. Together with $m^{q-1} \equiv 1 \pmod{q}$ by Fermat's Little Theorem, we have
$$d \mid q-1 \implies v_p(q-1) \ge v_p(d) = \alpha + 1.$$
Thus, any prime divisor $q$ of $n$ satisfies $q \equiv 1 \pmod{p^{\alpha+1}}$. Because $n$ is a product of these prime divisors, we deduce that $n \equiv 1 \pmod{p^{\alpha+1}}$. However, this contradicts to $v_p(n-1) = \alpha$. $\square$