Since $\left|2^{m}-5^{n}\right|$ is an odd number, and when $m=7, n=3$, $\left|2^{m}-5^{n}\right|=3$, if we can prove that when $n>1$, $\left|2^{m}-5^{n}\right| \neq 1$, then the minimum value sought is 3.

If there exist positive integers $m, n$, such that $n>1$, and $\left|2^{m}-5^{n}\right|=1$, then
$$2^{m}-5^{n}=1 \text{ or } 2^{m}-5^{n}=-1 \text{.}$$

If $2^{m}-5^{n}=1$, then $m \geqslant 3$. Taking both sides modulo 8, we require
$$5^{n} \equiv 7(\bmod 8)$$

But for any positive integer $n$, $5^{n} \equiv 1$ or $5(\bmod 8)$, which is a contradiction. Therefore, $2^{m}-5^{n}=1$ does not hold.
If $2^{m}-5^{n}=-1$, then by $n>1$, we know $m \geqslant 3$. Taking both sides modulo 8, we get
$$5^{n} \equiv 1(\bmod 8)$$

This implies that $n$ is even. Let $n=2x$, where $x$ is a positive integer, then
$$2^{m}=\left(5^{x}-1\right)\left(5^{x}+1\right)$$

Since $5^{x}-1$ and $5^{x}+1$ are two consecutive even numbers, this requires
$$5^{x}-1=2,5^{x}+1=4$$

which is impossible.
Therefore, the minimum value of $\left|2^{m}-5^{n}\right|$ is 3.