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I would like to know if there are "interesting" equivalences to the statement in the title. I am not interested in more general statements, like "every vector space has a Hamel basis" or "every vector space over (your favorite field) has a Hamel basis".

On the other hand, according to this post, such Hamel basis of $\mathbb R$ over $\mathbb Q$ may not exist, assuming the negation of AC. Where can I find a proof of this fact? Thanks.

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There is little known about this question (of equivalence).

We know that the statement implies (amongst other things) that there exists a non-measurable set. Therefore in a model where all sets are measurable such statement is false, i.e. there is no Hamel basis to $\mathbb R$ over $\mathbb Q$.

One can replace non-measurable by a weaker statement such as discontinuous solution to Cauchy's functional equation, or a set of real numbers without the Baire property. All such properties may fail without the axiom of choice (i.e. if ZFC is consistent then there are models in which the axiom of choice fails, and these properties also fail).

One of my favourite methods of proving the consistency of "There is no Hamel basis for $\mathbb R$ over $\mathbb Q$" is by means of automatic continuity. Namely there are models without choice in which whenever $T\colon V\to W$ is linear, $V$ is a Banach space and $W$ is a normed space, then $T$ is continuous. In particular $T$ is bounded.

Suppose that there was a Hamel basis to $\mathbb R$ over $\mathbb Q$, its cardinality would have to be $2^{\aleph_0}$, and therefore it has $2^{2^{\aleph_0}}$ many permutations, and each permutation of the basis would generate a linear function from $\mathbb R$ to itself.

However as a separable space $\mathbb R$ has only $2^{\aleph_0}$ many continuous endomorphisms, so in a model where automatic continuity for Banach spaces holds it is impossible that $\mathbb R$ has a Hamel basis over $\mathbb Q$.

Two important examples for such models are Solovay's model in which all sets of reals are Lebesgue measurable, although to obtain this model we need to assume the existence of an inaccessible cardinal. The second model is Shelah's model in which all sets of reals have the Baire property (and there are non-measurable sets), and for this model one only has to assume the consistency of ZFC -- without the additional axiom as in Solovay's case.

It is not known whether or not the existence of such basis implies that the real numbers can be well-ordered, which I presume is a main source of interest.

You can find a small diagram describing some of the known implications around this statement in Herrlich's The Axiom of Choice, Diagram 7.23 p. 156. In that book you could also find the wanted proof that $\mathbb R$ does not have a Hamel basis over $\mathbb Q$ in some models of ZF, Corollary 7.20 p. 154.

Also relevant:

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