Three, (50 points) Find all positive integer triples $(x, y, z)$, such that $z$ is a prime number, and $z^{x}=y^{3}+1$.
Three, Solution: $z^{x}=(y+1)\left(y^{2}-y+1\right)$, since $y$ is a positive integer, then $y+1 \geqslant 2$,
Let $y+1=z^{t}(t \in \mathbf{N}, 1 \leqslant t \leqslant x)$, then $y=z^{t}-1$, substitute into $y^{2}-y+1=z^{x-t}$, (1)
Obviously, $z^{x-t} \leqslant z^{2 t}$, i.e., $x \leqslant 3 t$, (1) simplifies to $z^{2 t}-z^{x-t}=3\left(z^{t}-1\right)$, then $z^{x-t}\left(z^{3 t-x}-1\right)=3\left(z^{t}-1\right)$.
(1) When $z=2, z^{3 t-x}-1$ and $z^{t}-1$ are both odd, so $z^{x-t}$ is also odd, thus $x=t$,
Therefore, $y^{2}-y+1=1$, so $y=1$, thus $x=1$, the solution set is $(1,1,2)$.
(2) When $z \neq 2, z^{3 t-x}-1$ and $z^{t}-1$ are both even, so $z^{x-t}$ is odd, since 3 is a prime, then
$3 \mid z^{x-t}$ or $3 \mid z^{3 t-x}-1$.
When $3 \mid z^{x-t}$, $3 \mid z$, so $z=3$, thus $x-t=1$, so $y^{2}-y+1=3$, solving gives $y=2, x=2$, the solution set is $(2,2,3)$. When $3 \mid z^{3 t-x}-1$, then $z^{x-t} \mid z^{t}-1$, thus $\left(z^{t}, z^{x-t}\right)=1$, then $x=t$, same as (1) solving gives $x=1$, contradiction.
In summary, all positive integer sets are $(1,1,2)$ or $(2,2,3)$.