Four, (50 points) Determine all positive integer triples $(x, y, z)$ such that $x^{3}-y^{3}=z^{2}$, where $y$ is a prime number, and $z$ is not divisible by 3 and $y$.

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Four, Solution: From the problem, we have $(x-y)\left[(x-y)^{2}+3 x y\right]=z^{2}$. Since $y$ is a prime number, and $z$ cannot be divisible by 3 and $y$, we get $(x, y)=1, (x-y, 3)=1$. Therefore, $\left(x^{2}+x y+y^{2}, x-y\right)=(3 x y, x-y)=1$.
Thus, we can set $x-y=m^{2}, x^{2}+x y+y^{2}=n^{2}, z=m n\left(m, n \in \mathbf{N}^{*}\right)$.
Hence, $3 y^{2}=4 n^{2}-(2 x+y)^{2}=(2 n-2 x-y)(2 n+2 x+y)$.
Since $y$ is a prime number, and $2 n-2 x-y<2 n+2 x+y$, there are only the following three cases:
(1) $2 n-2 x-y=y, 2 n+2 x+y=3 y$, which gives $x=0$, discarded.
(2) $2 n-2 x-y=3, 2 n+2 x+y=y^{2}$, thus
$y^{2}-3=4 x+2 y=4\left(m^{2}+y\right)+2 y=4 m^{2}+6 y$, i.e., $(y-3)^{2}-4 m^{2}=12$.
Solving this, we get $y=7, m=1$, so $x=8, y=7, z=13$.
(3) $2 n-2 x-y=1, 2 n+2 x+y=3 y^{2}$, thus $3 y^{2}-1=4 x+2 y=4\left(m^{2}+y\right)+2 y=2\left(2 m^{2}+3 y\right)$, i.e., $3 y^{2}-6 y-3 m^{2}=m^{2}+1$. Therefore, $m^{2}+1 \equiv 0(\bmod 3)$. But this contradicts $m^{2} \equiv 0,1(\bmod 3)$.
In summary, the only positive integer solution that satisfies the conditions is $(8,7,13)$.