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
${\mathfrak n},{\mathfrak l}<2^{\mathfrak m}$.
The $\aleph$-numbers are cardinalities which can be well ordered. Since each well-ordered set can be identified with an ordinal, each $\aleph$-number is the cardinality of some ordinal. If we consider the least ordinal of this cardinality we will have an initial ordinal (i.e. an ordinal which cannot be bijected with any ordinal strictly smaller than itself).
The axiom of choice is equivalent to the assertion that every set can be well ordered, therefore every infinite cardinality is an $\aleph$-number.
Suppose that the axiom of choice does not hold, then we have some sets which cannot be well-ordered. Cardinalities which cannot be well-ordered cannot be $\aleph$-numbers.
This gets worse. It is consistent that every set can be linearly ordered, and the axiom of choice still fails. It is even consistent that there is a set that cannot be linearly ordered at all (an amorphous set cannot be linearly ordered, for example). In Thomas Jech's book The Axiom of Choice he has a chapter about cardinal arithmetics without choice. It shows how wild the universe of ZF can be with respect to cardinals, when the axiom of choice is absent.
A few examples:
It is consistent that there is no "canonical" representative to the cardinalities. When the axiom of choice is absent, we are reduced to define cardinalities as equivalent classes in the relation of "there exists a bijection", by using Scott's trick we have that this is a well defined relation. Where as the axiom of choice allows us to pick a representative from each equivalence class (namely, initial ordinals) it is consistent that without the axiom of choice there is no way to choose such representative from all cardinalities at once. (i.e. there is no definable function which returns exactly one element of each cardinality)
We can have that for all infinite sets $|\{0,1\}\times A|=|A|$ but the axiom of choice still does not hold (this in contrast to $|A|\times|A| = |A|$ for every infinite $A$ implies choice)
The continuum hypothesis is independent of choice, that is there is a model in which the axiom of choice does not hold, and there are no sets of cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$.
These are just examples to show how cardinals can behave like a bunch of unsupervised teenagers in a house full of alcohol.
Best Answer
In what follows, by a "real", I mean a subset of $\omega\times\omega$, that is, a binary relation on $\omega$. (You can start with a bijection $\pi:\mathbb R\to\mathcal P(\omega\times\omega)$, which can be constructed without choice, so this is fine.)
If this relation happens to be a well-order of $\omega$ in order type $\omega+\alpha$, map the real to $\alpha$. Otherwise, map the real to $0$. This map is a surjection.
By the way, without choice, you cannot inject $\aleph_1$ into $\mathbb R$.