Does every number divide some $n^2+1$?

No.

$4$ cannot divide such an $n^2+1$, because modulo $4$, you have $n^2+1=1$ or $2$.

If $m$ divides $n^2+1$, then $n$ is a unit mod $m$ of order $4$. No prime of the form $4k+3$ can divide $n^2+1$ because in this case the group of units mod $p$ has order $4k+2$, which is not a multiple of $4$.

Let $p$ be an odd prime. Then $n^2\equiv -1 \bmod p$ is possible only if $-1$ is a quadratic residue modulo $p$.

This occurs precisely when $p\equiv 1 \bmod 4$ ie for $p=5, 13$ but not $p=3, 7, 11$

Now suppose $p$ is a factor of $m$ and that you want to find $n$ with $n^2+1=km$. Modulo $p$, this becomes $n^2\equiv -1$, so any odd prime factor of $m$ must be of the form $p=4r+1$.

There is more to do to specify precisely which $m$ are possible, but this is a start on the general problem.