$x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$. How do I prove that the diophantine equation $x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$.

We have
$$x^4-4y^4=z^2\iff\left(2y^2\right)^2+z^2=\left(x^2\right)^2.$$
Therefore if a primitive solution exists then $2y^2=2nm, \ z=m^2-n^2, \ x^2=m^2+n^2.\;$  [Note: Primitive here means one such that $x^2, 2y^2$ and $z$ are coprime which has as a consequence that $n$ and $m$ are coprime.]
From $2y^2=2nm$ and because $n$ and $m$ are coprime we get that both $n$ and $m$ are perfect squares and therefore $n=k^2,m=\ell^2$ for some $k$ and $\ell$.
This means that $x^2=\ell^4+k^4$, which has no integer solutions (see this).
