Let $V$ be a vector space over a finite field $F$.
Assume $V$ has a basis $S$. Then, define $\Phi:V^*\rightarrow F^S:f\mapsto f\upharpoonright S$. It can be shown that $\Phi$ is an isomorphism. Thus, $V^*\cong F^S$. Thus if $S'$ is another basis for $V$, then $F^S\cong F^{S'}$. Does this imply $S\approx S'$ in ZF?
It doesn't even imply $S\approx S'$ in ZFC.
Suppose that $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$. Consider now $S=\omega$ and $S'=\omega_1$. Both have duals of cardinality $\aleph_2$, and since the cardinality of a vector space is the $\max\{|F|,\dim V\}$, since $F$ is finite it has to be that both these duals have the same dimension: $\aleph_2$. Therefore they are isomorphic.
In the context of ZF, it is consistent to have a vector space with two bases of distinct cardinality.