There is a general argument without choice: Suppose ${\mathfrak m}+{\mathfrak m}={\mathfrak m}$, and ${\mathfrak m}+{\mathfrak n}=2^{\mathfrak m}$. Then ${\mathfrak n}=2^{\mathfrak m}.\,$ This gives the result.

The argument is part of a nice result
of Specker showing that if CH holds
for both a cardinal ${\mathfrak m}$
and its power set $2^{\mathfrak m}$,
then $2^{\mathfrak m}$ is
well-orderable. This shows that GCH
implies choice, and that the proof is
"local". It is still open whether CH
for ${\mathfrak m}$ implies that
${\mathfrak m}$ is well-orderable.

Anyway, here is the proof of the statement above: Note first that $2^{\mathfrak m}\cdot 2^{\mathfrak m}=2^{{\mathfrak m}+{\mathfrak m}}=2^{\mathfrak m}={\mathfrak m}+{\mathfrak n}$.

Let $X$ and $Y$ be disjoint sets with $|X|={\mathfrak m}$, $|Y|={\mathfrak n}$, and fix a bijection $f:{\mathcal P}(X)\times{\mathcal P}(X)\to X\cup Y$.

Note that there must be an $A\subseteq X$ such that the preimage $f^{-1}(X)$ misses the fiber $\{A\}\times {\mathcal P}(X)$. Otherwise, the map that to $a\in X$ assigns the unique $A\subseteq X$ such that $f^{-1}(a)$ is in $\{A\}\times {\mathcal P}(X)$ is onto, against Cantor's theorem.

But then, for any such $A$, letting $g(B)=f(A,B)$ gives us an injection of ${\mathcal P}(X)$ into $Y$, i.e., $2^{\mathfrak m}\le {\mathfrak n}$. Since the reverse inclusion also holds, we are done by Schroeder-Bernstein.

(Note the similarity to Apostolos's and Joriki's answers.)

The original reference for Specker's result is Ernst Specker, "Verallgemeinerte Kontinuumshypothese und Auswahlaxiom", Archiv der Mathematik 5 (1954), 332–337. A modern presentation is in Akihiro Kanamori, David Pincus, "Does GCH imply AC locally?", in "Paul Erdős and his mathematics, II (Budapest, 1999)", Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, (2002), 413–426.

Note that assuming that ${\mathfrak m}$ is infinite is not enough for the
result. For example, it is consistent
that there are infinite Dedekind
finite sets $X$ such that ${\mathcal P}(X)$ is also Dedekind finite. To be
Dedekind finite means that any proper
subset is strictly smaller. But if
$2^{\mathfrak m}$ is Dedekind finite
and $2^{\mathfrak m}={\mathfrak n}+{\mathfrak l}$ for nonzero
cardinals ${\mathfrak n},{\mathfrak l}$, then we must have
${\mathfrak n},{\mathfrak l}<2^{\mathfrak m}$.