How do I find solutions for $(x,y,p)$?

Question

I have the following equation-

$$p^x=y^4+4$$

Where $p$ is a prime and $x$ and $y$ are any natural numbers.

How do I find solutions for sets of $(p,x,y)$ satisfying the equation?I think congruence is to be applied but I am not being able to do it.

Clearly, $p=5,x=1$ and $y=1$ is a solution but is there a more systematic way to prove it rather than trial and error method.

Thanks for any help!!

Answer 1

HINT

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

By the Euclidean algorithim, $$\gcd(y^2-2y+2, y^2+2y+2)=\gcd(4y,y^2+2y+2)=1 \text{ or }2$$

If $\gcd(4y,y^2+2y+2)=1$, this implies $y^2-2y+2=1$.

Thus $y=1, p=5, x=1$ is a solution.

Can you take it from here?