I am happy to report that the equation has no solution. I kept my original response, and put the remaining arguments in the "EDIT" section below.

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

$$ \frac{\log m!}{3}>\log(n-3)\geq\sum_{p\equiv 2 \ (3), \ p\leq m}v_p(m!)\log p> \sum_{p\equiv 2\ (3), \ p\leq m} \left(\frac{m}{p}-1\right)\log p.$$

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 some bounds based on the work of Bordelles (see [here][1]) one can see that
$$\left|\sum_{p\leq m}\frac{\chi(p)\log p}{p}\right|<3\left|\frac{L'(1,\chi)}{L(1,\chi)}\right|+2.59<3.7,$$
where $\chi$ is the nontrivial character modulo $3$.
By including the contribution of the prime $p=3$ to $n-3$ in the original inequality, and using also the classical bounds of Rosser-Schoenfeld (1961), it follows that
$$\frac{m(\log m-0.9)}{3}>\frac{m(\log m-7.2)}{2}\quad\text{for}\quad m>e^{20}.$$
Hence in fact $m < e^{20}$, in which range I checked with SAGE that
$$\sum_{p\leq m}\frac{\chi(p)\log p}{p}<-0.04.$$ 
This inequality can be used to improve the previous bound to
$$\frac{m(\log m-0.99)}{3}>\frac{m(\log m-3.44)}{2}\quad\text{for}\quad m>e^{8.5},$$
which forces $m < e^{8.5}<5000$. In this range I checked with SAGE that the equation has no solution.

  [1]: http://www.emis.de/journals/JIPAM/article539.html?sid=539