Mike: If you fix an ordinal $\alpha$, then it is consistent that ${\mathfrak c}>\aleph_\alpha$. More precisely, there is a (forcing) extension of the universe of sets with the same cardinals where the inequality holds.
If you begin with a model of GCH, then you can go to an extension where ${\mathfrak c}=\aleph_\alpha$ and no cardinals are changed, as long as $\alpha$ is not a limit ordinal of countable cofinality. For example, $\aleph_{\aleph_\omega}$ is not a valid size for the continuum. But it can be larger.
Here, the cofinality of the limit ordinal $\alpha$ is the smallest $\beta$ such that there is an unbounded function $f:\beta\to\alpha$. There is a result of König that says that $\kappa^\lambda>\kappa$ if $\lambda$ is the cofinality of $\kappa$. If $\kappa={\mathfrak c}$, this says that $\lambda>\omega=\aleph_0$, since ${\mathfrak c}=2^{\aleph_0}$ and $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$. Since $\aleph_{\aleph_\omega}$ has cofinality $\omega$, it cannot be ${\mathfrak c}$.
But this is the only restriction! The technique to prove this (forcing) was invented by Paul Cohen and literally transformed the field.
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
${\mathfrak n},{\mathfrak l}<2^{\mathfrak m}$.
Best Answer
You are correct that without the axiom of choice $2^{\aleph_0}\newcommand{\CH}{\mathsf{CH}}$ may not be an $\aleph$. Therefore the continuum hypothesis split into two inequivalent statements:
Whereas the second variant implies that the continuum is well-ordered, the first one does not.
You suggested a third variant:
Let's see why $\CH_3\implies\CH_2\implies\CH_1$, and that none of the implications are reversible.
Note that if we assume $\CH_3$, then it has to be that $2^{\aleph_0}\leq\aleph_1$ and therefore must be equal to $\aleph_1$. If we assume that $\CH_2$ holds, then every cardinal less or equal to the continuum is finite or an $\aleph$, so $\CH_1$ holds as well.
On the other hand, there are models of $\sf ZF+\lnot AC$, such that $\CH_1$ holds and $\CH_2$ fails. For example, Solovay's model in which all sets are Lebesgue measurable is such model.
But $\CH_2$ does not imply $\CH_3$ either, because it is consistent that $2^{\aleph_0}=\aleph_1$, and there is some infinite Dedekind-finite set $X$, that is to say $\aleph_0\nleq |X|$. Therefore we have that $\aleph_0<|X|+\aleph_0$. Assuming $\CH_3$ would mean that if $X$ is infinite, then either $\aleph_0=|X|$ or $2^{\aleph_0}\leq|X|$. This is certainly false for infinite Dedekind-finite sets (one can make things stronger, and use sets that have no subset of size $\aleph_1$, while being Dedekind-infinite).
One can also think of the continuum hypothesis as a statement saying that the continuum is a certain kind of successor to $\aleph_0$. As luck would have it, there are $3$ types of successorship between cardinals in models of $\sf ZF$, and you can find the definitions in my answer here.
It is easy to see that $\CH_1$ states "$2^{\aleph_0}$ is a $1$-successor or $3$-successor of $\aleph_0$", and $\CH_3$ states that "$2^{\aleph_0}$ is a $2$-successor of $\aleph_0$" -- while not explicitly, it follows from the fact that I used to prove $\CH_3\implies\CH_2$.
So where does $\CH_2$ gets here? It doesn't exactly get here. Where $\CH_1$ and $\CH_3$ are statements about all cardinals, $\CH_2$ is a statement only about the cardinality of the continuum and $\aleph_1$. So in order to subsume it into the $i$-successor classification we need to add an assumption on the cardinals in the universe, for example every cardinal is comparable with $\aleph_1$ (which is really the statement "$\aleph_1$ is a $2$-successor of $\aleph_0$").
All in all, the continuum hypothesis can be phrased and stated in many different ways and not all of them are going to be equivalent in $\sf ZF$, or even in slightly stronger theories (e.g. $\sf ZF+AC_\omega$).
Without the axiom of choice we can have two notions of ordering on the cardinals, $\leq$ which is defined by injections and $\leq^*$ which is defined by surjections, that is to say, $A\leq^* B$ if there is a surjection from $B$ onto $A$, or if $A$ is empty. These notions are clearly the same when assuming the axiom of choice but often become different without it (often because we do not know if the equivalence of the two orders imply the axiom of choice, although evidence suggest it should -- all the models we know violate this).
So we can formulate $\CH$ in a few other ways. An important fact is that $\aleph_1\leq^*2^{\aleph_0}$ in $\sf ZF$, so we may formulate $\CH_4$ as $\aleph_2\not\leq^*2^{\aleph_0}$. This formulation fails in some models while $\CH_1$ holds, e.g. in models of the axiom of determinacy, as mentioned by Andres Caicedo in the comments.
On the other hand, it is quite easy to come up with models where $\CH_4$ holds, but all three formulations above fail. For example the first Cohen model has this property.
All in all, there are many many many ways to formulate $\CH$ in $\sf ZF$, which can end up being inequivalent without some form of the axiom of choice. I believe that the correct way is $\CH_1$, as it captures the essence of Cantor's question.
Interesting links: