$x^2+y^5 = z^2$

Assuming we are talking about real numbers $x$ , $y$ , and $z$ . If we fix $y=0$ , then $x^2 = z^2$ , $\implies x =\pm z$ , therefore, there are infinitely many solutions . Of course there will be infinitely many solutions if $x,y, z \in \mathbb C$ .

$x^2+y^5=z^2$ $y^5=z^2-x^2$ $y^5=(z+x)(z-x)$

If $z$ is a number such that $z=x+1$ , then $z+x=2x+1$ and $z-x=1$ . Therefore: $y^5=2x+1$

For every odd $y$ , there is some $x$ that fulfills $y^5=2x+1$ , because $y^5$ is odd if $y$ is odd, and $2x+1$ can cover every odd integer number. Therefore, there are an infinite number of solutions.