[Solution] Transform the original equation into
$$y^{3}=\left(z^{2}-x\right)\left(z^{2}+x\right) .$$

Since $y$ is a prime number, we get
$$\left\{\begin{array}{l} z^{2}-x=1 \\ z^{2}+x=y^{3} \end{array}\right.$$

or $\left\{\begin{array}{l}z^{2}-x=y, \\ z^{2}+x=y^{2}\end{array}\right.$.
From (1), we get $x=(z-1)(z+1)$,
thus $z-1=1, z=2$,
hence $x=3, y^{3}=7$,
which contradicts the fact that $y$ is a prime number, so the system of equations (1) has no prime number solutions.
From (2), we get $x=(y-z)(y+z)$,
thus $y-z=1, y=z+1$,
hence $x=2 z+1$.
Substituting into the first equation of (2) gives
$$z^{2}-3 z-2=0$$

This equation has no prime number solutions, so the system of equations (2) has no prime number solutions.
In conclusion, the original equation has no prime number solutions.