10. Find all positive integer triples $(x, y, z)$ such that $y$ is a prime number, 3 and $y$ are not divisors of $z$, and $x^{3}-y^{3}=z^{2}$.
10. From the original equation, we get $z^{2}=(x-y)\left(x^{2}+x y+y^{2}\right)$. Let
$$\begin{aligned}
\left(x-y, x^{2}+x y+y^{2}\right) & =\left(x-y,(x-y)^{2}+3 x y\right)=(x-y, 3 x y) \\
& =\left(x-y, 3 y^{2}\right)=d
\end{aligned}$$

Then $d \mid 3 y^{2}$. Since $y$ is a prime and 3 and $y$ are not factors of $z$, we have $d=1$. Thus, $x-y$ and $x^{2}+x y+y^{2}$ are both perfect squares. We set $x-y=u^{2}, x^{2}+x y+y^{2}=v^{2}$, where $u, v \in \mathbf{N}^{*}$. Multiplying both sides of the equation $x^{2}+x y+y^{2}=v^{2}$ by 4, completing the square, rearranging, and factoring, we get
$$(2 v-2 x-y)(2 v+2 x+y)=3 y^{2}$$

Noting that $y$ is a prime, $2 v+2 x+y \in \mathbf{N}^{*}$, and $2 v-2 x-y<2 v+2 x+y$, we can only have
$$\left\{\begin{array}{l}
2 v-2 x-y=1, y \text { or } 3, \\
2 v+2 x+y=3 y^{2}, 3 y \text { or } y^{2} .
\end{array}\right.$$

For the first case, we have
$$3 y^{2}-1=2(2 x+y)=4 u^{2}+6 y$$

Thus, $u^{2}+1 \equiv 0(\bmod 3)$. But $u^{2} \equiv 0$ or $1(\bmod 3)$, so there is no solution in this case.
For the second case, we have
$$2 y=4 x+2 y$$

Leading to $x=0$, which also has no solution.
For the third case, we have
$$y^{2}-3=4 u^{2}+6 y$$

That is, $(y-2 u-3)(y+2 u-3)=12$, solving which gives $(y, u)=(7,1)$, and thus $(x, z)=(8,13)$.

The solution to the original equation that satisfies the conditions is $(x, y, z)=(8,7,13)$.