Prime

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

The two terms $y^2+2y+2$ and $y^2-2y+2$ have at least two different primes and there will some composite terms because

$gcd(y^2+2y+2,y^2-2y+2)=gcd(y^2-2y+2,4y)$ .As $y$ is odd implies $y^2-2y+2$ is odd.So, gcd can be one or some other values.

If the gcd is $1$ then these two factors can not share a prime then $y^2-2y+2=1 => y=1$ so $x=1 ,p=5$ .

if gcd is other than $1$ i.e. $p^r$ for some positive integer $r$ then $p^r | y$ so $y=p^r.k$ and the factor becomes $y^2+2y+2=k^2.p^{2r}-2.k.p^r+2$ which is not divisible by $p^r$

So, the only solution is $p=5,x=1,y=1$