I understand!! @DonLarynx I have also an other question...

We have that $x$ and $z$ are odd and $y$ is even and that $x^4+y^4=z^2 \Rightarrow y^4=(z-x^2)(z+x^2)$.

$(x, y, z)=1$

To show that $gcd(z-x^2, z+x^2)=2$, is the following the only way??

Let $(z-x^2, z+x^2)=d>1$. Then it has a prime divisor, let $p$.

$p \mid d , d \mid z-x^2 \Rightarrow p \mid z-x^2$

$p \mid d , d \mid z+x^2 \Rightarrow p \mid z+x^2$

So $p \mid 2x^2 \Rightarrow p \mid 2 \text{ OR } p \mid x$
and $p \mid 2z \Rightarrow p \mid 2 \text{ OR } p \mid z$