(I believe this material is in Jech's giant set theory book, but I don't have it in front of me right now.)

I am almost entirely certain that the existence of a non-measurable set of reals is not equivalent to the full axiom of choice, but rather to some weakened choice principle. It certainly follows from the well-orderability of $\mathbb{R}$; I suspect it requires substantially less, but I am not sure.

[Meta-edit: Below I originally had a paragraph of bad explanation of why a nonmeasurable set doesn't imply choice. I've replaced it with a better explanation.]

EDIT: Here's why full choice is overkill. Take a model of ZFC (so there is a set of reals which is nonmeasurable), and consider a forcing over this model which is countably closed, so it adds no new reals. In particular, the ground model had a well-ordering of the reals, and since no new reals are added, and well-orderedness is absolute between forcing extensions, the reals in the extension are still well-ordered.

Now take a symmetric submodel of the generic extension by this forcing in which choice fails. Since the reals in the extension are well-orderd, we can still build a non-measurable set. In essence, what we've done is to add a failure of choice "way high up" in the cumulative hierarchy, where it doesn't affect sets of reals. The resulting model will be a model of $ZF+\neg AC+$``There exists a non-measurable set of reals," and so the existence of a non-measurable set of reals can't be equivalent to the full AC.

A natural question at this point is to ask whether a nonmeasurable set implies the axiom of choice for $\mathbb{R}$; the answer is still no, for pretty much the same reason. In general, a purely existential property (``there is a non-measurable set") which only talks about objects of a bounded rank in the cumulative hierarchy has no chance of implying full choice.

However, it is true that the existence of a non-measurable set requires some amount of the axiom of choice: we can build models of ZF+ `Every subset of $\mathbb{R}$ is measurable," so the existence of a nonmeasurable set requires _some_ amount of choice. Even this statement, though, is not completely straightforward, since it implies the consistency of ZFC. So one thing we can ask is what the consistency strength of`

ZF doesn't prove that there is a non-measurable set" is.

Certainly the axiom of determinacy implies that every set of reals is measurable, but this is overkill in terms of consistency strength. The most famous model in which every set of reals is measurable is called the Solovay model, and is a model of ZF+DC (so something more than dependent choice is required to prove the existence of a non-measurable set). The construction of the Solovay model assumed an inaccessible cardinal (much weaker than what is needed for AD), and Saharon Shelah later proved that the consistency of one inaccessible cardinal is implied by the consistency of "every set of reals is measurable."

So the current situation is a dichotomy: either inaccessible cardinals are inconsistent with ZFC, or some amount of choice greater than dependent choice is required to ensure the existence of a non-measurable set. Given that much much larger large cardinals are generally thought to be consistent with ZFC, the consensus is that ZF does not prove the existence of a non-measurable set of reals. However, I do not know of any reasonably natural choice principle which is exactly equivalent to the existence of a non-measurable set. I hope this answers your question!