I conjecture that there exist infinitely many integers $n$ such that $$(n^{2015}+1)\mid n!.$$
I have seen a simpler problem that there exist infinitely many integers $n$ such that $(n^2+1)\mid n!$.
Alternatively, I considered the Pell equation $n^2+1=5m^2$, $2m<n$, but for $2015$ I can't figure it out.
On
Since $2015 = 5\cdot 13\cdot 31 $, and $n^a + 1| n^{ab}+1 $ if $b$ is odd, a necessary condition for $n^{2015}+1 | n! $ is $n^m+1 | n!$ for every $m$ in $\{5, 13, 31 , 5\cdot 13 , 5\cdot 31 , 13\cdot 31 \} $.
Solutions are going to be hard to find. All those expressions of the form $n^j-n^{j-1}+...-n+1 $ for odd $j$ will have to have all prime factors $\le n$ in order to divide $n!$.
On
It suffices to show that for infinitely many $n$, the largest prime factor of $n^{2015}+1$ is at most $\sqrt{n}$. Indeed, if $n$ is such a large integer and $p$ is a prime, then the largest value of $a$ for which $p^a\mid n^{2015}+1$ is $\leq c \log n$ for some constant $c$, while $n!$ is divisible by $p^a$ with $a\geq \frac{n}{p}-1\geq \sqrt{n}-1>c \log n$. It was shown by Schinzel (Theorem 13 in https://eudml.org/doc/204826) that for any nonzero integers $A$ and $B$, any integer $k\geq 2$ and any $\varepsilon>0$ there exist infinitely many integers $n$ such that the largest prime factor of $An^k+B$ is less than $n^{\varepsilon}$. In particular, the claim of the problem holds with $2015$ replaced by any positive integer.
Modest progress. There are infinitely many integers $n$ such that $n^3+1\mid n!$.
We always have $n^3+1=(n+1)(n^2-n+1)$. Let $n=k^2+1$. Then $$ n^2-n+1=(1+k+k^2)(1-k+k^2). $$ Assume further that $k\equiv1\pmod3$. In that case $1+k+k^2$ and $n+1=2+k^2$ are both divisible by $3$. For all sufficiently large $k\equiv1\pmod3$ we thus have $$ (k^2+1)^3+1=3^2\cdot\frac{k^2+2}3\cdot\frac{k^2+k+1}3(k^2-k+1) $$ that is clearly a factor of $(k^2+1)!$.