The claim is correct. We begin with a standard lemma.

**Lemma:** Let $p$ be an odd prime, and $x$ an integer coprime to $p$. Let $k$ be a positive integer and let $o$ be the multiplicative order of $x$ modulo $p$. Then $$v_p(x^k-1) = \begin{cases} v_p(x^o-1) + v_p(\frac{k}{o}) & o \mid k \\ 0 & o \nmid k \end{cases}.$$

We prove the claim by comparing the $p$-adic valuation of $P_n(m)$ and $n!$ for every prime $p \le n$ (since all prime divisors of $n!$ are $\le n$). We skip the primes in the set $\{ q: q \mid m \} \cup \{2\}$ (they may be treated separately). The lemma implies that your claim is equivalent to the following inequalities:
$$(*) \forall p\le n (\text{odd, coprime to }m): v_p((\frac{n}{o})!) + \lfloor \frac{n}{o} \rfloor v_p(m^o-1) \ge v_p(n!),$$
$$\text{where }o \text{ is the multiplicative order of m modulo p},$$
and slightly different inequalities for $p=2$ and $p \mid m$.

We have:
$$v_p((\frac{n}{o})!) + \lfloor \frac{n}{o} \rfloor v_p(m^o-1) \ge v_p((\frac{n}{o})!) + \lfloor \frac{n}{o} \rfloor = \sum_{i \ge 1} \lfloor \frac{n}{op^i} \rfloor + \lfloor \frac{n}{o} \rfloor$$
$$= \sum_{i \ge 1} \lfloor \frac{pn}{op^i} \rfloor \ge \sum_{i \ge 1} \lfloor \frac{n}{p^i} \rfloor = v_p(n!), $$
where the last inequality follows from the fact that $o<p$. This proves $(*)$.

To deal with the exceptional primes ($2$ and divisors of $m$), just note that $P_n(m)$ is divisible by $m^{\binom{n}{2}} (m-1)^n$ and that $v_p(n!)\le n,\binom{n}{2}$ for every $p$ and $n$.