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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. Assume that $m\geq 1000$, and denote by $\chi$ the nontrivial character modulo $3$. Then $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<\sum_{n\leq m}\frac{\chi(n)\Lambda(n)}{n}-\sum_{p\leq m,\ p\neq 3}\frac{\log p}{p^2+p}. $$ This implies, in combination with some ideas of Bordelles (cf. the proof of (4.2) here), that $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<3\left|\frac{L'(1,\chi)}{L(1,\chi)}\right|+1.53<2.64\ .$$ By including the contribution of the prime $p=3$ to $n-3$ in the original inequality, and using also some classical bounds by Rosser and Schoenfeld (cf. (3.15) and (3.21) here), it follows that $$\frac{m(\log m-0.9)}{3}>\frac{m(\log m-6.1)}{2}\quad\text{for}\quad m>e^{16.5}.$$ Hence $m < e^{16.5}$. I checked with SAGE that in fact $$\sum_{p\leq m}\frac{\chi(p)\log p}{p}<-0.63\quad\text{for}\quad e^{16.5}>m>e^{7}.$$ This can be used to improve the previous bound to $$\frac{m(\log m-0.99)}{3}>\frac{m(\log m-2.92)}{2}\quad\text{for}\quad e^{16.5}>m>e^{7},$$ which in turn forces $m < 1000$. The above shows that all solutions of the original equation satisfy $m < 1000$. However, I checked with SAGE that in this range the equation has no solution.

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. Assume that $m\geq 1000$, and denote by $\chi$ the nontrivial character modulo $3$. Then $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<\sum_{n\leq m}\frac{\chi(n)\Lambda(n)}{n}-\sum_{p\leq m,\ p\neq 3}\frac{\log p}{p^2+p}. $$ This implies, in combination with some ideas of Bordelles (cf. the proof of (4.2) here), that $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<3\left|\frac{L'(1,\chi)}{L(1,\chi)}\right|+1.53<2.64\ .$$ By including the contribution of the prime $p=3$ to $n-3$ in the original inequality, and using also some classical bounds by Rosser and Schoenfeld (cf. (3.15) and (3.21) here), it follows that $$\frac{m(\log m-0.9)}{3}>\frac{m(\log m-6.1)}{2}\quad\text{for}\quad m>e^{16.5}.$$ Hence $m < e^{16.5}$. I checked with SAGE that in fact $$\sum_{p\leq m}\frac{\chi(p)\log p}{p}<-0.63\quad\text{for}\quad e^{16.5}>m>e^{7}.$$ This can be used to improve the previous bound to $$\frac{m(\log m-0.99)}{3}>\frac{m(\log m-2.92)}{2}\quad\text{for}\quad e^{16.5}>m>e^{7},$$ which in turn forces $m < 1000$. The above shows that all solutions of the original equation satisfy $m < 1000$. However, I checked with SAGE that in this range the equation has no solution.

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. Assume that $m\geq 1000$, and denote by $\chi$ the nontrivial character modulo $3$. Then $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<\sum_{n\leq m}\frac{\chi(n)\Lambda(n)}{n}-\sum_{p\leq m,\ p\neq 3}\frac{\log p}{p^2+p}. $$ This implies, in combination with some ideas of Bordelles (cf. the proof of Lemma 4.1 here), that $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<3\left|\frac{L'(1,\chi)}{L(1,\chi)}\right|+1.53<2.64\ .$$ By including the contribution of the prime $p=3$ to $n-3$ in the original inequality, and using also some classical bounds by Rosser and Schoenfeld (cf. (3.15) and (3.21) here), it follows that $$\frac{m(\log m-0.9)}{3}>\frac{m(\log m-6.1)}{2}\quad\text{for}\quad m>e^{16.5}.$$ Hence $m < e^{16.5}$. I checked with SAGE that in fact $$\sum_{p\leq m}\frac{\chi(p)\log p}{p}<-0.63\quad\text{for}\quad e^{16.5}>m>e^{7}.$$ This can be used to improve the previous bound to $$\frac{m(\log m-0.99)}{3}>\frac{m(\log m-2.92)}{2}\quad\text{for}\quad e^{16.5}>m>e^{7},$$ which in turn forces $m < 1000$. The above shows that all solutions of the original equation satisfy $m < 1000$. However, I checked with SAGE that in this range the equation has no solution.

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. Assume that $m\geq 1000$, and denote by $\chi$ the nontrivial character modulo $3$. Then $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<\sum_{n\leq m}\frac{\chi(n)\Lambda(n)}{n}-\sum_{p\leq m,\ p\neq 3}\frac{\log p}{p^2+p}. $$ This implies, in combination with some ideas of Bordelles (cf. the proof of (4.2) here), that $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<3\left|\frac{L'(1,\chi)}{L(1,\chi)}\right|+1.53<2.64\ .$$ By including the contribution of the prime $p=3$ to $n-3$ in the original inequality, and using also some classical bounds by Rosser and Schoenfeld (cf. (3.15) and (3.21) here), it follows that $$\frac{m(\log m-0.9)}{3}>\frac{m(\log m-6.1)}{2}\quad\text{for}\quad m>e^{16.5}.$$ Hence $m < e^{16.5}$. I checked with SAGE that in fact $$\sum_{p\leq m}\frac{\chi(p)\log p}{p}<-0.63\quad\text{for}\quad e^{16.5}>m>e^{7}.$$ This can be used to improve the previous bound to $$\frac{m(\log m-0.99)}{3}>\frac{m(\log m-2.92)}{2}\quad\text{for}\quad e^{16.5}>m>e^{7},$$ which in turn forces $m < 1000$. The above shows that all solutions of the original equation satisfy $m < 1000$. However, I checked with SAGE that in this range the equation has no solution.

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) 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.

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. Assume that $m\geq 1000$, and denote by $\chi$ the nontrivial character modulo $3$. Then $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<\sum_{n\leq m}\frac{\chi(n)\Lambda(n)}{n}-\sum_{p\leq m,\ p\neq 3}\frac{\log p}{p^2+p}. $$ This implies, in combination with some ideas of Bordelles (cf. the proof of Lemma 4.1 here), that $$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<3\left|\frac{L'(1,\chi)}{L(1,\chi)}\right|+1.53<2.64\ .$$ By including the contribution of the prime $p=3$ to $n-3$ in the original inequality, and using also some classical bounds by Rosser and Schoenfeld (cf. (3.15) and (3.21) here), it follows that $$\frac{m(\log m-0.9)}{3}>\frac{m(\log m-6.1)}{2}\quad\text{for}\quad m>e^{16.5}.$$ Hence $m < e^{16.5}$. I checked with SAGE that in fact $$\sum_{p\leq m}\frac{\chi(p)\log p}{p}<-0.63\quad\text{for}\quad e^{16.5}>m>e^{7}.$$ This can be used to improve the previous bound to $$\frac{m(\log m-0.99)}{3}>\frac{m(\log m-2.92)}{2}\quad\text{for}\quad e^{16.5}>m>e^{7},$$ which in turn forces $m < 1000$. The above shows that all solutions of the original equation satisfy $m < 1000$. However, I checked with SAGE that in this range the equation has no solution.