Logarithms?

$Let \: m \in \mathbb{N}$

$5^p+4p^4=m^2 \Rightarrow 5^p+(2p^2)^2=m^2 \Rightarrow 5^p=m^2-(2p^2)^2 \Rightarrow 5^p=(m+2p^2)(m-2p^2)$

Case 1: $5 \vert (m+2p^2),(m-2p^2) \Rightarrow 5 \vert (m+2p^2)-(m-2p^2) \Rightarrow 5 \vert 4p^2 \Rightarrow 5 \vert p \Rightarrow p=5$

$p=5 \Rightarrow 5^p+4p^4=5625=75^2$

Case 2:

$(m-2p^2)=1,(m+2p^2)=5^p \Rightarrow 5^p-1=4p^2 \Rightarrow p \vert (5^p-1)$

Using FLT and the fact that $p \neq 5$ we have:

$5^p-1 \equiv 5-1 \equiv 4 \pmod p \\ p \vert (5^p-1) \Rightarrow p \vert 4 \Rightarrow p=2$

$p=2 \Rightarrow 5^p+4p^4=89 \neq m^2$

So our only solution is $\boxed{p=5}$ .