If $\mathscr F$ is an ultrafilter on an infinite set $M$, then it can be shown that $|\mathscr F|=2^{|M|}$.

We know that for each subset $A\subseteq M$ we have $A\in\mathscr F$ or $M\setminus A \in \mathscr F$. In this way we can divide the power set $\mathscr P(M)$ into two disjoint parts $\{A\subseteq M; A\in\mathscr F\}$ and $\{A\subseteq M; M\setminus A\in\mathscr F\}$, both of them have cardinality $|\mathscr F|$.

Thus we have $$|\mathscr F|=|\mathscr F|+|\mathscr F|=|\mathscr P(M)|=2^{|M|}.$$

Here we have used that $a+a=a$ for infinite cardinals; which uses axiom of choice. (At least the proof of this fact based on $a\le a+a \le a\cdot a=a$ uses AC.)

Is there an argument showing $|\mathscr F|=2^{|M|}$ without using AC?

(I am aware that already the existence of free ultrafilters can be considered a weaker form of AC, but I am not sure whether it is sufficiently strong to imply $a+a=a$. Additionally, the above argument makes sense for principal ultrafilters, too.)