Example 6 Let $M \subseteq\{1,2, \cdots, 2011\}$ satisfy: in any three elements of $M$, there can always be found two elements $a, b$ such that $a \mid b$ or $b \mid a$. Find the maximum value of $|M|$. ${ }^{[2]}$
(2011, China Western Mathematical Olympiad)
Solve for when
$$
M=\left\{2^{k}, 3 \times 2^{l} \mid k=0,1, \cdots, 10 ; l=0,1, \cdots, 9\right\}
$$

the condition is satisfied, at this time, $|M|=21$.
Assume $|M| \geqslant 22$, let the elements of $M$ be
$$
a_{1}2011$, a contradiction.
In summary, $|M|_{\max }=21$.