Does bounded Zermelo construct any cumulative hierarchy? ZF is sufficient to construct the von Neumann hierarchy, and prove that every set appears at some stage $V_\alpha$. This is the basis for Scott's trick, for instance. But how much of ZF is needed? Is bounded Zermelo/Mac Lane set theory enough, no Choice assumed? I know Foundation is necessary, and I'm not getting rid of that. I've seen something called the "rank axiom" in discussion of second-order version of original Zermelo (these notes), but I'm sure people have finely calibrated what precisely is needed.
To be honest, all I really want is an ordinal-valued rank function such that sets of rank at most $\alpha$ form a set, for all $\alpha$, and all sets have a rank. So if the von Neumann hierarchy doesn't work, I'm happy to work with something else (and for 'ordinals', I don't need von Neumann ordinals).
 A: One can directly assume in Zermelo that every set belongs to a rank.  This does not add any strength at all.  But notice that Zermelo may not prove the existence of very many von Neumann ordinals:  there may be well-orderings which do not have a von Neumann ordinal as order type.  A natural model of this theory is the union
of the V^{omega + n}'s for n a natural number.  Notice that in this structure the axioms of Zermelo hold, every object belongs to a V_alpha, but omega+omega does not exist.  There are well-orderings with order types far higher than omega+omega (and in this context the Scott representation of the ordinals is available).
There are a couple of additional remarks.  It is interesting to note
that the assertion that every set has a rank adds no strength at all to
Zermelo set theory (or to Zermelo set theory with bounded separation) but that adding this assertion to KP, a theory much weaker than Zermelo, gives a theory much stronger than Zermelo.  The reason is that KP has a lot of replacement.
The natural way to describe the rank function in "Zermelo with ranks" is probably to use the Scott ordinals as values of the rank function but note that the rank function is not necessarily onto the ordinals.  In the absence of replacement, the von Neumann notion of ordinal simply isn't the right notion of ordinal number.
A: KP (Kripke-Platek set theory) is the most well-known fragment of $\sf{ZF}$ which suffices for the development of the rank function, thus $\sf{KPR}$ = $\sf{KP}$ + "for all ordinals $\alpha$,  $V(\alpha)$ exists" is the usual minimal theory in which one can be assured of the stratification of the universe into $V(\alpha)$s.
On the other hand, as observed by Mathias, $\sf{KPR}$ proves that Zermelo set theory $\sf{Z}$ has a transitive model, so in particular, $\sf{KPR}$ proves that $\sf{Z}$ is consistent; see Lemma 6.31 of this preprint, which was later published in APAL (2001).
Therefore, by the second incompleteness theorem, even Zermelo set theory (let alone bounded Zermelo set theory) cannot interpret $\sf{KPR}$.
Finally, I will add that $\sf{KPR}$ is provable in the well-known extension $\sf{KP}^{\cal{P}}$ of $\sf{KP}$, which is also studied in Mathias' paper.
