Calm down, it is easy

Let $P$ be a prime so:

Case 1 : If $P=5$

$4P^2+1=101$ and $6P^2+1=151$ which are both primes so $5$ is a solution

Case 2 : If $P \equiv 1,2,3,4 \pmod{5}$

So $P^2 \equiv 1,or,-1 \pmod{5}$

if $P^2 \equiv 1 \pmod{5}$ so $4P^2+1 \equiv 0 \pmod{5}$ which is not prime.

if $P^2 \equiv -1 \pmod{5}$ so $6P^2+1 \equiv 0 \pmod{5}$ which is not prime

So $5$ is the only solution.