Here is a quick proof that there are only finitely many solutions.
We use $m!=(n-3)(n^2+3n+9)$. Here $n$ is divisible by $3$, hence $n^2+3n+9$ is not divisible by any prime $p\equiv 2\pmod{3}$. In other words, all the prime divisors $p\equiv 2\pmod{3}$ of $m!$ are contained in $n-3$ with multiplicity. It follows, with the usual notations, that
$$ \frac{\log m!}{3}>\log(n-3)\geq\sum_{p\equiv 2 \ (3)}v_p(m!)\log p> \sum_{p\equiv 2\ (3), \ p\leq m} \left(\frac{m}{p}-1\right)\log p.$$
The left hand side is $\sim (m\log m)/3$, while the right hand side is $\sim (m\log m)/2$ by Dirichlet's theorem. Hence for large $m$ the inequality must fail.
EDIT. Using a crude bound based on the work of Bordelles I can see that $$\left|\sum_{p\leq m}\frac{\chi(p)\log p}{p}\right|<10,$$ where $\chi$ is the nontrivial character modulo $3$. Hence for $m>100$ the original bound implies $$\frac{m\log m}{3}>\frac{m(\log m-13)}{2},$$ so that $m < e^{39}<10^{17}$.