4. (Bulgaria 2000) Find all positive integer triples $(x, y, z)$ such that $z$ is not divisible by 3 and $y$, and $y$ is a prime number, and $x^{3}-y^{3}=z^{2}$.
4. From $x^{3}-y^{3}=z^{2}$, factorizing both sides we get: $(x-y)\left[(x-y)^{2}+3 x y\right]=z^{2}$

Since $y$ is a prime number, and $y \backslash z$, it follows that $(x, y)=1$, so $(x-y, y)=1,(x-y, x)=1$.
Furthermore, $(x-y, 3)=1$, so $\left((x-y),(x-y)^{2}+3 x y\right)=1$.
Let $x-y=m^{2}(1),(x-y)^{2}+3 x y=n^{2}$ (2), $z=m n$ (3).
Multiplying both sides of (2) by 4 we get: $4 x^{2}+4 x y+4 y^{2}=4 n^{2}$, which is $3 y^{2}=(2 n)^{2}-(2 x+y)^{2}=(2 n+2 x+y)$ $(2 n-2 x-y)$.
$1^{0}\left\{\begin{array}{l}2 n-2 x-y=1 \\ 2 n+2 x+y=3 y^{2}\end{array} \Rightarrow 3 y^{2}-1=4 x+2 y=4 m^{2}+6 y \Rightarrow 3(y-1)^{2}-4=4 m^{2}\right.$.
Taking modulo 3 on both sides, the above equation clearly has no solution.
$$
\begin{array}{c}
2^{0}\left\{\begin{array}{l}
2 n-2 x-y=3 \\
2 n+2 x+y=y^{2}
\end{array} \Rightarrow y^{2}-3=4 x+2 y=4 m^{2}+6 y \Rightarrow(y-3)^{2}-12=4 m^{2} \Rightarrow(y-3+2 m)\right. \\
(y-3-2 m)=12 \Rightarrow y-3+2 m=6, y-3-2 m=2 \Rightarrow y=7, m=1, x=8, z=13 .
\end{array}
$$
$3^{0}\left\{\begin{array}{l}2 n-2 x-y=y^{2} \\ 2 n+2 x+y=3\end{array} \Rightarrow x=0, n=0, z=0\right.$ which contradicts $x, y, z \in \mathbf{N}^{+}$.
In conclusion, $y=7, m=1, x=8, z=13$.