This question arose in a discussion with Peter Doyle.

It is provable in ZF that one can divide by any positive finite cardinal $k$: if $X \times \{1,\ldots,k\} \simeq Y \times \{1,\ldots,k\}$ then $X \simeq Y,$ where I use $\simeq$ to denote the existence of a bijection between two sets.

Assuming the Axiom of Choice, the positive finite cardinals are precisely the cardinals that one can divide by since $1\cdot\mathfrak{m} = 2\cdot\mathfrak{m}$ for any infinite cardinal $\mathfrak{m}$. The ultimate question is whether this is actually provable in ZF:

If $\mathfrak{m}$ is an infinite cardinal, must there be two cardinals $\mathfrak{p} \neq \mathfrak{q}$ such that $\mathfrak{p}\cdot\mathfrak{m} = \mathfrak{q}\cdot\mathfrak{m}$?

As a candidate for a counterexample, I thought it might be possible to divide by the cardinal of a (strongly) amorphous set. Specifically, in the First Fraenkel model:

Is it true that if $X \times A \simeq Y \times A$ then $X \simeq Y$, where $A$ is the set of atoms?