What is it with Primes and Squares?

$5^p+4p^4=x^2 \implies 5^p=x^2- 4p^4=(x^2-2p^2)(x^2+2p^2)$

$x^2-2p^2=5^{t} \ , \ x^2+2p^2=5^{p-t} \ , \ \lfloor \frac{p}{2} \rfloor \geq t \geq 0$

$2p^2+2p^2=5^{p-t}-5^t \implies 4p^2=5^{p-t}-5^t \implies p^2=5^t \frac{5^{p-2t}-1}{4}$

$t$ can be either $\{0,1,2\}$ , for $p^2$ is a power two of a prime.

If $t=2$ then $\frac{5^{p-4}-1}{4}=1 \implies 5^{p-4}=5 \implies p=5$

If $t=1$ then $\frac{5^{p-2}-1}{4}=5 \implies 5^{p-2}=21$ and no $p$ satisfies here.

If $t=0$ then $5^{p}-1=4p^2$ . one can check that for $p \geq 2$ we always have $5^{p}-1 > 4p^2$ (note that $5^p$ increases with greater rate, compared to $p^2$ ), so no $p$ passes here either.