The set $\{-1, 3, 5, -2, 6, 10\}$ attests that $M$ can have $6$ elements: The sum of any two numbers from the triplet $(-1, 3, 5)$ is a power of two and the same is true for triplet $(-2, 6, 10)$. For the sake of contradiction, assume that some set $M$ has more than $6$ elements.

Clearly, $M$ can't contain three (or more) non-positive numbers. Hence it contains at least five positive numbers. Denote by $x$ the largest positive number in $M$ and by $a, b, c, d$ some four other positive numbers in $M$. Consider pairs $x + a, x + b, x + c, x + d$. They are all larger than $x$ and less than $2x$. The open interval $(x, 2x)$ contains at most one power of two, hence at least three of the four sums are not a power of two. Without loss of generality, assume those are $x + a, x + b, x + c$. Considering the triplets $(a, b, x), (a, c, x), (b, c, x)$ we infer that all $a + b, a + c, b + c$ are powers of two. However, this is impossible. Without loss of generality, let $a = \max\{a, b, c\}$. Then $a + b$ and $a + c$ both lie in $(a, 2a)$, hence at least one of them is not a power of two, a contradiction.