Example 7 Let $p$ be a given positive integer, try to determine the minimum positive value of $(2 p)^{2 m}-(2 p-1)^{n}$, where $m, n$ are any positive integers.
The smallest positive value sought is $(2 p)^{2}-(2 p-1)^{2}=4 p-1$. To prove this, we first note that by

and
$$\begin{array}{c}
(2 p)^{2}=(4 p-2) p+2 p \\
(2 p-1)^{2}=(4 p-2)(p-1)+(2 p-1)
\end{array}$$

it is easy to deduce that
$$(2 p)^{2 m}-(2 p-1)^{n} \equiv(2 p)-(2 p-1) \equiv 1(\bmod 4 p-2)$$

Furthermore, we prove that there are no positive integers $m, n$ such that $(2 p)^{2 m}-(2 p-1)^{n}=1$. Suppose the contrary, then we have
$$\left((2 p)^{m}-1\right)\left((2 p)^{m}+1\right)=(2 p-1)^{n}$$

The two factors on the left side are clearly coprime, while the right side is the $n$-th power of a positive integer, hence
$$(2 p)^{m}+1=a^{n}$$

where $a$ is a positive integer, and $a \mid(2 p-1)$. Taking (2) modulo $a$, the left side becomes
$$(2 p-1+1)^{m}+1 \equiv 1+1 \equiv 2(\bmod a)$$

leading to $2 \equiv 0(\bmod a)$, but since $a \mid 2 p-1$, $a$ is an odd number greater than 1, which leads to a contradiction.
Combining (1), if $(2 p)^{2 m}-(2 p-1)^{n}>0$, then $(2 p)^{2 m}-(2 p-1)^{n} \geqslant 4 p-1$, and equality is achieved when $m=1, n=2$, thus proving our conclusion.