Is it natural?

First note that if $mn-1|n^2+1$ then $mn-1|m^2n^2-1+n^2+1=n^2(m^2+1)$ It follows that $mn-1|m^2+1$ , so if $(m,n)$ is an answer $(n,m)$ will be too.

Let $n=1$ , it gives $m=2,3$ , we get 4 pairs from here: $(1,2),(2,1),(1,3),(3,1)$ Note that if $\frac{n^2+1}{mn-1}=r$ then $r\equiv-1 \ \ \text{mod} \ n$ , and $r$ have the form of $kn-1$ for some positive integer $k$ . There is no answer for $m=n$ therefore we can suppose $m>n$ and we can write $kn-1=\frac{n^2+1}{mn-1}<\frac{n^2+1}{n^2-1}=1+\frac{2}{n^2-1}<2$ which follows that $kn<3$ and $n=2$ this gives us $m=3$ . So we get 2 other solutions from here: $(2,3),(3,2)$ Therefore there are 6 such pairs.