II. (50 points) Determine all positive integer triples $(x, y, z)$ such that $x^{3}-y^{3}=z^{2}$, where $y$ is a prime, $y \chi_{z}, 3 \nmid z$.

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From the problem, we have
$$
(x-y)\left[(x-y)^{2}+3 x y\right]=z^{2} \text {. }
$$

Since $y$ is a prime number, and $y \gamma_{z}, 3 Y_{z}$, combining with equation (1), we know
$$
(x, y)=1,(x-y, 3)=1 \text {. }
$$

Then $\left(x^{2}+x y+y^{2}, x-y\right)=(3 x y ; x-y)=1$.
From equations (1) and (2), we get
$$
x-y=m^{2}, x^{2}+x y+y^{2}=n^{2}, z=m n
$$
$\left(m, n \in \mathbf{N}_{+}\right)$.
Thus, $3 y^{2}=4 n^{2}-(2 x+y)^{2}$
$$
=(2 n+2 x+y)(2 n-2 x-y) \text {. }
$$

Since $y$ is a prime number, and $2 n-2 x-y<2 n+2 x+y$, there are the following 3 cases:
(i) $2 n-2 x-y=y, 2 n+2 x+y=3 y$.
This gives $x=0$, which is discarded.
(ii) $2 n-2 x-y=3,2 n+2 x+y=y^{2}$. Therefore,
$$
y^{2}-3=4 x+2 y=4\left(m^{2}+y\right)+2 y=4 m^{2}+6 y \text {, }
$$

which simplifies to $(y-3)^{2}-4 m^{2}=12$.
Solving this, we get $y=7, m=1$.
Thus, $x=8, y=7, z=13$.
(iii) $2 n-2 x-y=1,2 n+2 x+y=3 y^{2}$. Therefore,
$$
\begin{array}{l}
3 y^{2}-1=4 x+2 y=4\left(m^{2}+y\right)+2 y \\
=2\left(2 m^{2}+3 y\right),
\end{array}
$$

which simplifies to $3 y^{2}-6 y-3 m^{2}=m^{2}+1$.
Thus, $m^{2}+1 \equiv 0(\bmod 3)$.
But this contradicts $m^{2} \equiv 0,1(\bmod 3)$.
In summary, the only positive integer solution that satisfies the conditions is $(8,7,13)$.