I've heard it said that theorems based on choice are also available in ZF "a few powersets away", and I think this is one of them, but I'm not sure how to prove it. (I'm also interested to hear of other formulas of this type. A special case: If $A\not\prec\omega$, then $\omega\preceq{\cal PP}(A)$, i.e. the double powerset of a Dedekind-finite set is infinite. This one I can prove though.)
Edit: Just to be clear, we say $A\preceq B$ if there exists an injection $f:A\to B$, and $A\prec B$ if $A\preceq B$ and $B \mathrel{\diagup\hskip{-1em}\preceq} A$.
To make the statement precise, what I am asking is if there exists an $n\in\omega$ such that
$${\sf ZF}\vdash \forall A,B[A\prec B\vee B\prec{\cal P}^n(A)],$$
and if so, what is the smallest such $n$. Andres' comment below suggests another, much weaker generalization: is the statement $\exists\alpha\in{\sf On}\,\forall A,B\,(A\prec B\vee B\prec{\cal P}^\alpha(A))$ provable? Here ${\cal P}^\alpha(A)$ is defined by transfinite recursion: ${\cal P}^0(A)=A$, ${\cal P}^{\alpha+1}(A)={\cal P(P}^\alpha(A))$, and ${\cal P}^\alpha(A)=\bigcup_{\beta<\alpha}{\cal P}^\beta(A)$ for limit ordinals $\alpha$.
The answer is negative in $\sf ZF+\lnot AC$.
Let $A$ be a set which cannot be well-ordered, pick $\alpha\in\sf Ord$, and let $B=\aleph(\mathcal P^\alpha(A))$, then we have that $A$ and $B$ are incomparable, but also $\mathcal P^\alpha(A)$ and $B$ are incomparable.