What are all the integral solutions of $n!=m(m^2-1)$?

In Richard Guy's third edition of Unsolved Problems in Number Theory, he writes on p.301 in D25 Equations involving factorial $n$. that

Simmons notes that $n!=(m-1)m(m+1)$ for $(m,n)=(2,3),(3,4),(5,5),$ and $(9,6)$ and asks if there are other solutions. More generally he asks if there are any other solutions of $n!+x=x^k.$

This Simmons is in reference to:

Gustavus J. Simmons, A factorial conjecture, J. Recreational Math., 1(1968) 38.

So it would seem that this problem was open in 2004 when Guy wrote the third edition of his book, but I don't know if it has been solved in the last 17 years.

Interestingly enough, there might be something even deeper going on. Let us define a family of elliptic curves given by the Weierstrass equations

$$E_n: y^2=x^3-x-n!$$

Your question is asking whether $E_n$ has any integer solutions $(x,0)$, which by the Nagell-Lutz theorem is equivalent to asking if $E_n$ has any points of 2-torsion. A natural generalization of this conjecture is to ask whether or not the torsion group $E_n(\mathbb{Q})$ is trivial for any $n\neq3,4,5,6$.

Using Sage one can compute the torsion groups of elliptic curves, and computationally it checks out that $E_n(\mathbb{Q})$ is trivial for any $n\leq10,000$, except for $n=3,4,5,6$ where we have that the only torsion points were the points of 2-torsion.

It turns out, in fact, that this generalized problem is $\mathbf{equivalent}$ to your conjecture. This is true because $x^3-x$ is always a multiple of $3$ (check by hand of by Fermat's Little Theorem) so any non $2$-torsion solution $y^2=x^3-x-n!$ will also be a multiple of $3$. By Nagell Lutz this would imply that the discriminant $\Delta_{E_n/\mathbb{Q}}=-4(27n!^2-4)$ would be a multiple of $3$ which as we can clearly see it is not.