I'm aware that the statement "for all cardinals $\kappa$, $\kappa^2 = \kappa$" is equivalent to the axiom of choice (I believe this was proved by Tarski).
More generally, does anyone know if the statement "for all cardinals $\lambda, \kappa$, $\lambda^2 \leq \kappa^2$ implies $\lambda \leq \kappa$" is equivalent to AC?
It is the case that if $\kappa^2=\lambda^2\implies\kappa=\lambda$ (where $\kappa$ and $\lambda$ are arbitrary cardinals, not just $\aleph$ numbers), then the axiom of choice holds. (See the proof here: Does $A\times A\cong B\times B$ imply $A\cong B$?)
Note that the statement you suggest implies this, since $\kappa^2=\lambda^2$ implies both $\leq$ and $\geq$.
