Characterize all $m \in \mathbb N$ such that $\frac{m^2-1}{m^2-n} \in \mathbb N$ and $1<n<m^2-1$ This just came when I was considering when $m^2-1$ can be divisible by more than two numbers $1 \leq n \leq m^2-1$.

Since $n=1$ and $n=m^2-1$ are always solutions the question is about when it is possible to have $$\dfrac{m^2-1}{m^2-n} \in \mathbb N$$ for $1&lt;n&lt;m^2-1$

How to characterize all such $m$?
There are solutions for all $m &gt; 2$: simply let $n = m^2 - m + 1$, then
$$
\frac{m^2 - 1}{m^2 - n} = \frac{m^2 - 1}{m - 1} = m + 1.
$$
Essentially your question is: when is $m^2 - 1$ prime? And the answer is: essentially never, because $m^2 - 1 = (m-1)(m+1)$, and unless $m = 2$ (or the unusual case $m = 1$) this gives two non-trivial factors.
To start with $$\frac{m^2-1}{m^2-n}-1=\frac{n-1}{m^2-n}$$
So $$n\lt m^2\lt 2n$$

For the first $10000$ values of $n$, 8537 have no solutions for $m$ in this range, 
1244 have one solution, 170 have two, 48 have three and one, namely 4621,
has four solutions, in this case $m=68,69,71,76$.  All have $m=1$ as well.  $4621$ also has a negative solution for $m=67$.