Three. (25 points) Find all positive integer triples $(x, y, z)$ such that $1+2^{x} \times 3^{y}=5^{z}$ holds.
(Zhang Lei)
Three, taking the original equation modulo 3 yields that $z$ is even.
Let $z=2 r\left(r \in \mathbf{N}_{+}\right)$. Then
$\left(5^{r}-1\right)\left(5^{r}+1\right)=2^{x} \times 3^{y}$.
Since $\left(5^{\prime}-1,5^{\prime}+1\right)=\left(5^{\prime}-1,2\right)=2$, we have
$\left(\frac{5^{r}-1}{2}, \frac{5^{\prime}+1}{2}\right)=1$.
Thus, $\frac{5^{\prime}-1}{2} \cdot \frac{5^{\prime}+1}{2}=2^{x-2} \times 3^{y}(x \geqslant 2)$.
Also, $\frac{5^{r}-1}{2}$ is even, and $\frac{5^{r}+1}{2} \geqslant 3$, so
$\frac{5^{r}-1}{2}=2^{x-2}, \frac{5^{r}+1}{2}=3^{y}$.
Hence, $3^{y}-2^{x-2}=1$.
When $x=2$, $y$ is not an integer.
When $x=3$, $y=1, z=2$.
When $x \geqslant 4$, taking equation (1) modulo 4 shows that $y$ is even.
Let $y=2 t\left(t \in \mathbf{N}_{+}\right)$. Then
$\left(3^{t}-1\right)\left(3^{t}+1\right)=2^{x-2}$.
Since $\left(3^{t}-1,3^{t}+1\right)=\left(3^{t}-1,2\right)=2$, we have
$\left(\frac{3^{t}-1}{2}, \frac{3^{t}+1}{2}\right)=1$.
Therefore, $\frac{3^{t}-1}{2} \cdot \frac{3^{t}+1}{2}=2^{x-4}(x \geqslant 4)$.
Since $\frac{3^{t}-1}{2}<\frac{3^{t}+1}{2}$, we have $\frac{3^{t}-1}{2}=1$.
Solving gives $t=1, y=2$.
From $\frac{5^{\prime}+1}{2}=3^{y}$, we know that $r$ is not an integer, so the original equation has no positive integer solutions in this case.
In summary, the positive integer solution to the original equation is $(x, y, z)=(3,1,2)$.