One, (40 points) Find the smallest positive integer $m$ such that the equation
$$
2^{x}+3^{y}-5^{z}=2 m
$$

has no positive integer solutions for $x, y, z$.
$$
\begin{array}{l}
2=2^{2}+3-5,4=2+3^{3}-5^{2}, \\
6=2^{3}+3-5,8=2^{2}+3^{2}-5, \\
10=2^{3}+3^{3}-5^{2}, 12=2^{3}+3^{2}-5, \\
14=2^{4}+3-5,16=2^{5}+3^{2}-5^{2}, \\
18=2^{4}+3^{3}-5^{2}, 20=2^{4}+3^{2}-5 .
\end{array}
$$

Below we prove that $2^{x}+3^{y}-5^{z}=22$ has no solution.
If $x=1$, then $3^{y}=20+5^{z}$.
Clearly, this equation has no solution.
When $x \geqslant 2$, $2^{x} \equiv 0(\bmod 4)$, then $(-1)^{y} \equiv 3(\bmod 4)$.
Thus, $y$ is odd.
$$
\text { Also, } 2^{x}+3^{y}-5^{z} \equiv(-1)^{x}-(-1)^{z} \equiv 1(\bmod 3) \text {, }
$$

Therefore, $x$ is odd and $z$ is even.
Let $x=2 \alpha+1, y=2 \beta+1\left(\beta \in \mathbf{N}, \alpha \in \mathbf{N}_{+}\right)$.
Then $2^{x}+3^{y}-5^{z}=2^{2 \alpha+1}+3^{2 \beta+1}-5^{z}$
$=2 \times 4^{\alpha}+3 \times 9^{\beta}-5^{z}$
$\equiv 2(-1)^{\alpha}+3(-1)^{\beta}$
$\equiv 0$ or 1 or $-1(\bmod 5)$.
But $22 \equiv 2(\bmod 5)$, so this equation has no integer solution.
Therefore, the smallest positive integer $m$ that satisfies the condition is 11.