One can check that $M = \{1, 2, 2^2, 2^3, \dots, 2^{10}, 3, 3 \times 2, 3 \times 2^2, \dots, 3 \times 2^9\}$ satisfies the condition, and $|M| = 21$.

Suppose that $|M| \ge 22$, and let $a_1 < a_2 < \dots < a_k$ be the elements of $M$, where $|M| = k \ge 22$. We first prove that $a_{n+2} \ge 2a_n$ for all $n$; otherwise, we have $a_n < a_{n+1} < a_{n+2} < 2a_n$ for some $n < k + 2$, then any two of these three integers $a_n, a_{n+1}, a_{n+2}$ do not have any multiple relationship, which contradicts the assumption.

It follows from the inequality above that $a_4 \ge 2a_2 \ge 4$, $a_6 \ge 2a_4 \ge 8$, $\dots$, $a_{22} \ge 2a_{20} \ge 2^{11} > 2011$, which is a contradiction!

Hence, the maximum value of $|M|$ is $21$.
$\square$