I hope to show that $x^4+5y^4=z^2$ doesn't have an integer solution.
You may guess that you can solve it using the infinite descent procedure.
I tried it but I had a trouble in solving it.
What I did :
Observe that we can say $(x,y)$=1.
$$-x^4+z^2=5y^4$$
$$-(x^2-z)(x^2+z)=5y^4$$
Set $x^2-z=u$ and $x^2+z=v$.
But since we don't know $(x,z)$, we cannot tell that $(u,v)=1,2$, so failed.
Another :
I observed that
$$k(x^2+ay^2)^2+k(x^2-ay^2)^2=rz^2$$
Then we get $2k=r$, $2ka^2=5r$, $a^2=5$, so failed because $a$ is not an integer.
If $x=t,y=2t,z=9t^2$ then $x^4+5y^4=81t^4=z^2.$
The parameter $t$ here just serves to remind that if $(x,y,z)$ is a solution so is $(tx,ty,t^2z)$. So one has families of solutions, the generator of the above being $(1,2,9)$. If one were to look for all solutions, finding the generators would be the task.
Update: I searched using maple for $x,y \le 1000$ and found the generator $(1,2,9)$ above along with another generator $(79,36,6881).$ Likely someone who knows more about elliptic curves than I do could show there are only finitely many different families (distinct generators with pairwise gcd 1) and maybe even find all such generators.