Sizes of bases of vector spaces without the axiom of choice

Yes, ZF+BPIT implies that vector space dimension is well-defined. [Edit: some Googling shows that James Halpern gave the same answer back in the 1960s.]

Working in ZF+BPIT, fix a field $F$ and an $F$-vector space $V$ and bases $A$ and $B$ of V. That is, each element of $V$ is a unique $F$-linear combination of elements of $A$; likewise for $B$. For each $a\in A$, let $S_a$ be the minimal subset of $B$ such that $a$ is spanned by $S_a$. Each $S_a$ is finite; give it the discrete topology. Let $X=\prod_{a\in A}S_a$, which is nonempty by BPIT (and is compact Hausdorff). By Schroeder-Bernstein, it suffices to show that some $f\in X$ is injective. By compactness, it suffices to show that for every finite subset $K$ of $A$, there is an $f\in X$ that is injective on $K$. Since each $\prod_{a\in A\setminus K}S_a$ is nonempty by BPIT, it suffices to show that there is an injection in every $\prod_{a\in K}S_a$. That is a nice little linear algebra exercise you can solve in ZF using the finite case of Hall's marriage theorem.