Are there any pairs of natural numbers $(m, n)$, with $5 \leq m$, other than $(5, 15128)$ and $(6, 366776)$, that satisfy the condition $(m! + 2)\sigma(n) = 2n \cdot m!$, where $\sigma(n)$ denotes the sum of divisors of $n$ and $m!$ denotes the factorial of $m$?
This question arises from the theory of immaculate groups (or, equivalently, Leinster groups). An immaculate group is a group, such that its order is equal to the sum of all orders of its proper normal subgroups.
It is easy to see, that if $A$ is a non-abelian simple group then $A\times\mathbb{Z}_n$ is immaculate iff $(|A|+1)\sigma(n) = 2|A|n$. Two well known examples of immaculate groups of that form are $A_5\times\mathbb{Z}_{15128}$ and $A_6\times\mathbb{Z}_{366776}$. In terms of immaculate groups this question thus can be reworded as: "Does there exist any immaculate group of the type $A_n \times \mathbb{Z}_n$ other than those two?".
I checked this condition for all $n \leq 10000$ and $5 \leq m \leq 7$ but found nothing.
Any help will be appreciated.
A similar question about $M_{11}$: Are there any natural numbers $n$ that satisfy the condition $7921\sigma(n) = 15840n$?
The equation can be written as $$ \Bigl(\frac{m!}{2}+1\Bigr)\sigma(n)=n\,m!. $$ Since $m!$ and $m!/2+1$ are coprime, $n$ is a multiple of $m!/2+1$. I have looked for solutions of the equation $$ \sigma\Bigl(k\,\Bigl(\frac{m!}{2}+1\Bigr)\Bigr)=k\,m! $$ for $5\le k\le20$, $1\le k\le10^6$. This search provided so far a new solution: $$ m=10,\quad n=691\,816\,586\,092 $$