Example 3 Let $M=\left\{1,2, \cdots, 2^{m} n\right\}\left(m, n\right.$ be positive integers) be a set of $2^{m} n$ consecutive positive integers. Find the minimum value of the positive integer $k$ such that any $k$-element subset of $M$ contains $m+1$ numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ satisfying $a_{i}$ divides $a_{i+1}(i=1,2, \cdots, m)$.
Take a $\left(2^{m}-1\right) n$-element subset $S_{0}=\left\{n+1, n+2, \cdots, 2^{m} n\right\}$ of $M$. From the previous analysis, we know that $S_{0}$ does not meet the conditions of the problem, so the minimum value of $k$ is $\geqslant\left(2^{m}-1\right) n+1$.

Next, let $S$ be any $\left(2^{m}-1\right) n+1$-element subset of $M$. We need to prove that there exist $m+1$ numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in $S$ such that $a_{i} \mid a_{i+1}(i=1,2, \cdots, m)$. For this, we construct the following $\left[\frac{n+1}{2}\right]$ drawers:
$$
A_{i}=\left\{(2 i-1) \cdot 2^{r} \mid r=0,1,2, \cdots \text {, and }(2 i-1) \cdot 2^{r} \in M\right\}\left(i=1,2, \cdots,\left[\frac{n+1}{2}\right]\right),
$$
so in $A_{i}$ there are exactly $m$ numbers in the interval $\left(n, 2^{m} n\right)$. In fact, for $1 \leqslant i \leqslant\left[\frac{n+1}{2}\right]$, there exists a unique positive integer $r_{0}$ such that $2^{r_{0}-1}(2 i-1) \leqslant n<2^{r_{0}}(2 i-1)$, thus
$$
\begin{aligned}
2^{r_{0}-1}(2 i-1) & \leqslant n<2^{r_{0}}(2 i-1)<2^{r_{0}+1}(2 i-1)<\cdots<2^{r_{0}+(m-1)}(2 i-1) \\
& =2^{m} \cdot 2^{r_{0}-1}(2 i-1) \leqslant 2^{m} \cdot n<2^{r_{0}+m}(2 i-1) .
\end{aligned}
$$

It is clear that $A_{i}$ contains exactly $m$ numbers $2^{r_{0}}(2 i-1), 2^{r_{0}+1}(2 i-1), \cdots, 2^{r_{0}+m-1}(2 i-1)$ in the interval $(n, \left.2^{m} n\right]$, so $A=A_{1} \cup A_{2} \cup \cdots A\left[\frac{n+1}{2}\right]$ contains a total of $m\left[\frac{n+1}{2}\right]$ numbers greater than $n$, and $\{1,2, \cdots, n\} \subseteq A$, so the number of elements in $A$ is $|A|=m\left[\frac{n+1}{2}\right]+n$, and the number of elements in $M \backslash A$ is $2^{m} n-m\left[\frac{n+1}{2}\right]-n$, thus $S$ contains at least
$$
\left(2^{n}-1\right) n+1-\left(2^{m} n-m\left[\frac{n+1}{2}\right]-n\right)=m\left[\frac{n+1}{2}\right]+1
$$

elements in $A$, so by the pigeonhole principle, $S$ contains at least
$$
\left[\frac{\left(m\left[\frac{n+1}{2}\right]+1\right)-1}{\left[\frac{n+1}{2}\right]}\right]+1=m+1
$$

elements $a_{1}, a_{2}, \cdots, a_{m+1}\left(a_{1}<a_{2}<\cdots<a_{m+1}\right)$ in the same subset $A_{i}$ $\left(1 \leqslant i \leqslant\left[\frac{n+1}{2}\right]\right)$, thus satisfying $a_{i} \mid a_{i+1}(i=1,2, \cdots, m+1)$.
In summary, the minimum value of $k$ is $\left(2^{m}-1\right) n+1$.