When $p=5$, both $4p^2+1=101$ and $6p^2+1=151$ are prime. Therefore $p=5$ is one solution. We are going to show that this is the only solution.

When $p\equiv \pm1 \pmod{5}$, we have $4p^2 + 1\equiv 0\pmod{5}$. This means that $(4p^2+1)$ is a multiple of $5$ which cannot be prime.

Meanwhile, when $p \equiv \pm 2 \pmod{5}$, we have $6p^2 +1\equiv 0\pmod{5}$. This means that $(6p^2+1)$ is a multiple of $5$ which is not prime.