How do I prove that there is infinite set of numbers $m$ such that the biggest prime divisor of $m^4+1$ is bigger than $2m$?
Let $p$ be a prime congruent to 1 modulo 8 --- there are infinitely many such. For each such prime, there are 8 numbers $a$, $0\lt a\lt p$, such that $p$ divides $a^8-1=(a^4-1)(a^4+1)$. So there are four numbers $m$ such that $p$ divides $m^4+1$. They come in pairs that add up to $p$, so two of them are less than $p/2$. Either one is thus an $m$ with a prime divisor exceeding $2m$.
Each $m$ has only finitely many $p$ dividing $m^4+1$, so there are infinitely many such $m$.