Almost same as IMO 2008: Let $p=8t+1$; we know there exist two solutions of $x^2+1 \equiv 0 \pmod p$, let $n$ be the smaller one. Then it holds $k=p-2n>0$ and $k$ is odd. We have $p|(p-2n)^2+4=k^2+4$, so $k^2+4 \ge p$. We must have $k^2+4>p$ because $p=8t+1$. Also it can't be $k^2+4=2p$ because $k$ is odd. So $k^2+4 \ge 3p>6n$; and this is enough when $p$ is big enough. Note: we also have $k^2+4 \not = 3p$ and $k^2+4 \not = 4p$ so there exist infinitely $n$ such $n^2+1$ has a prime bigger than $2n+ \sqrt{10n}$