Numbers disturbing

$\begin{aligned} n+1 &\mid n^2 + 1\\ \implies n+1 &\mid n^2+1 - (n+1)(n-1)\\ \implies n+1 &\mid n^2+1 - n^2 + 1\\ \implies n+1 &\mid 2\\ \implies n&= 1\\ \end{aligned}$

There is only one such positive integer: n = 1. In fact, n^2+l = n(n+l)-(n-l); thus, if n+1 l n^2+1, then n+1 l n-1 which for positive integer n is possible only if n-l = 0, hence if n = 1.