Diophantine equation: $x^4+4=py^4$ 
find all primes p for which $x^4+4=py^4$ is solvable for integers.

My try: I started out with $p=2$ . LHS  must be even, thus $(x^4+4)\mod(16)=4$ similarly
$(2y^4)\mod (16)=${$0,2$}. Thus for $p=2$ there are no solutions.
Now I tried factoring i.e $x^4+4=(x^2+2x+2)(x^2-2x+2)=py^4.$ I dont know what to do next
Any ideas??
 A: It is easy to see that $x$ is odd (otherwise $4$ divides $py^4$ and $x, y$ are both even, which gives contradiction mod $16$). Then the $\gcd$ of $x^2 + 2x + 2$ and $x^2 - 2x + 2$ is $1$. Therefore we must have \begin{eqnarray}x^2 + 2x + 2 &=& u^4\\x^2 - 2x + 2 &=& pv^4\end{eqnarray} or \begin{eqnarray}x^2 + 2x + 2 &=& pu^4\\x^2 - 2x + 2 &=& v^4.\end{eqnarray} In the first case, we have $(x^2 + 1)^2 + 1 = (u^2)^2$ which is impossible. In the second case, we have $(x^2 - 1)^2 + 1 = (v^2)^2$ which leads to $x = 1$, and hence $y = 1$ and $p = 5$.
A: @WhatsUp has the answer, but I wanted to show just for fun how to systematically derive the other case $(X,Y,P)=(0,\pm 1,4)$. It doesn't answer the question because $P=4$ is not prime, but to those who ask "How can we derive the other solution without some kind of trial and error?" well here it is.

Consider the equation $$(p^2-q^2)^2+(2pq)^2=(p^2+q^2)^2$$
Let $pq=4$ so that $$(p^2-q^2)^2+64=(p^2+q^2)^2$$
Let $p=x^2+y^2$ and $q=x^2-y^2$ so that $p^2-q^2=(2xy)^2$ and $p^2+q^2=2(x^2+y^2)^2$ then $$(2xy)^4+64=4(x^2+y^2)^4$$ or $$(xy)^4+4=\frac 14(x^2+y^2)^4$$
The coefficient is rational, but we want it to be an integer. The requirement is that $x^2+y^2$ is even. So let $x=2s$ and $y=2t$ so it follows $$(4st)^4+4=4(s^2+t^2)^4.$$
Remember the constraint $pq=4$. Since $p=x^2+y^2$ and $q=x^2-y^2$ then it follows $(s^2+t^2)(s^2-t^2)=1$. Since $x$ and $y$ are integers then $s$ and $t$ are integers and so $s^2\pm t^2$ are also integers. Therefore $s^2+t^2=s^2-t^2=1$ or $2s^2=2$ or $s=\pm 1$ and thus $t=0$.
Therefore we derive $(X,Y,P)=(0,\pm 1, 4)$.
