2018/014) Find all triples (p,x,y) such that $p^x= y^4+4$ where p is a prime and x and y are natural numbers

We have $p^x=(y^4+4) = (y^4+ 4y^2+4) - 4y^2= (y^2+2)^2-(2y)^2$
Or $p^x= (y^2+2y+2)(y^2-2y+2)$
Now as p is prime we have both $y^2+2y+2)$ and $(y^2-2y+2)$ are powers of p and as
$(y^2-2y+2) < (y^2+2y+2)$
So $(y^2-2y+2)$ divides $(y^2+2y+2)$ divides the difference that is 4y
So $y^2 - 2y +2 -4y <=0$ or $y^2-6y+2<=0$ or $(y-3)^2< 7$ or $ y - 3 < 3$ or $y < 6$
Putting y = 1 to 5 in original equation we see that (p,x,y) = (5,1,1) is the only solution