Let show that the equation in question is equivalent to another one:$x^4+y^4=z^2$, called as $\Gamma$-equation.
If $p$ and $q$ and $r$ satisfy $(\Gamma)$, then, upon setting $x=r$, $y=pq$, and $z=p^4-q^4$, we obtain a solution of $x^4=4y^4+z^2$. Conversely, if $x$, $y$, and $z$ satisfy the equation in question, then there are $m$ and $n$ such that $x^2=m^2+n^2$, $2y^2=2mn$, and $z=m^2-n^2$. Hence we find that there are $p$ and $q$ with $m=p^2$ and $n=q^2$. Then $x^2=p^4+q^4$. Therefore the two equations are indeed equivalent, speaking in integers.
Now the transformed equation $(\Gamma)$ is easy to show its insolvability, and here we offer two methods:
I. Here have I proved that there is no pythagorean triple $(a,b,c)$ with both $a$ and $b$ squares, nor both $a$ and $c$ squares. If $(\Gamma)$ has a solution $(x,y,z)$ then $(x^2,y^2,z)$ is a pythagorean triple contradiction what is just said. So there is no solution at all.
II. Here is a proof contained in Diophantine Equation by Mordell.
Suppose $(x,y,z)$ is a solution with $(x,y)=1$. Since $x$ and $y$ cannot both be odd by considerations of $2$-powers, we assume that $y$ is even, and $x$ is odd. Now by Euclid's lemma, we know that there are $a$ and $b$ such that $x^2=a^2-b^2$, $y^2=2ab$, and $z=a^2+b^2$, with $a, b$ coprime. By dint of the same lemma again, we find $p$ and $q$ such that $x=p^2-q^2$, $b=2pq$, $a=p^2+q^2$, with $p, q$ coprime. Since $p$, $q$ and $p^2+q^2$ are pairwise prime to each other, and for $y^2=2ab=4pq(p^2+q^2)$, there are coprime $r, s$ and $t$ such that $p=r^2$, $q=s^2$, and $p^2+q^2=t^2$. So $r^4+s^4=t^2$, while $z=a^2+b^2=p^4+q^4+6p^2q^2=r^8+s^8+6r^4s^4$. Hence $z>(r^4+s^4)^2=t^4$, namely, $t$ is strictly less than $z$. So this completes the proof "by descent".
Though the length is still a pain, I hope that, this time, length brings not only pain, but the result, the richness of the subject, and its depth, as well. And tell me if something occurs that is inappropriate here. Thanks in advance.