Prove that $5$ is the only prime $p$ such that $3p + 1$ is a perfect square

Prove that $5$ is the only prime $p$ such that $3p + 1$ is a perfect square

Prove that $5$ is the only prime $p$ such that $3p + 1$ is a perfect square.
I started off with assuming that $p$ is odd (since $2$ clearly does not
satisfy). This would mean that $3p + 1$ is even. Since if a perfect square
is even, it has to be divisible by four, it would mean that $4|3p + 1$.
If $p \equiv 1\space (mod \space 4)$, $3p + 1 \equiv 0\space (mod \space
4)$. All other possible residues of $p \space (mod \space 4)$ don't work
out. This implies that our proof can be reduced to finding all $m$ such
that $3(4m + 1) + 1 = (2n)^2$, or $3m + 1 = n^2$, which looks a lot like
our first expression, except, here $m$ does not have to be a prime.
It just remains to be proven that the only $m$ such that $4m + 1$ is a
prime and $3m + 1$ is a perfect square is $1$. I don't know where to go
from here.