I've tested each $0<n<10^5$, it seems that 7 is the only solution for $n!+1 \equiv 0 \mod 10n+1$. But I can't prove it correct or wrong.
I know that no solution exists if $10n+1$ is composite, because any factor p (p>1) of $10n+1$ must be greater than n, so $10n+1 \ge (n+1)^2$, that's $n \le 8$. I've checked them one by one, no solution found.
But when $10n+1$ is a prime, I can't work it out. Anyone can help?
Updated at Nov 26, 2021
Still no answer yet, just put some of my studies here.
When $n \gt 5$,
$n!+1 \equiv 0 (\mod 10n+1)$
$n!+1 \equiv 1 (\mod 10n)$
since $10n+1$ and $10n$ are coprime, Chinese remainder theorem gives $n!+1=(10n+1)^2 + 10n(10n+1)k$, $k \in \mathbb{Z}_{\ge 0} $.
If $k=0$, $n!+1=(10n+1)^2$, $n=7$ is the only solution.
If $k \gt 0$, no progress at all :(