Is every finitely generated idempotent ring singly generated as a two-sided ideal? In this post, a ring is understood to be what one usually calls a ring, not assuming that it has a unit. Some people call such objects rng.

Question: Let R be a finitely generated (non-unital and associative) ring, such that $R=R^2$, i.e. the multiplication map $R \otimes R \to R$ is surjective (every element is a sum of products of other elements). Is it possible that every element of $R$ is contained in a proper two-sided ideal of $R$? Or, must it be the case that $R$ is singly generated as a two-sided ideal in itself?

Note, if $Z \subset R$, then the ideal generated by $Z$ is the span of $Z \cup RZ \cup ZR \cup RZR$, which in the case of idempotent rings is equal to the span of $RZR$.
More generally, one can ask:

Question: For a fixed natural number $k$, can it happen that every set of $k$ elements of $R$ generates a proper ideal of $R$?

So far, I do not know of any example where the ring $R$ is not generated by a single element as a two-sided ideal in itself. I first thought that it must be easy to find counterexamples, but I learned from Narutaka Ozawa that the free non-unital ring on a finite number of idempotents is singly generated as a two-sided ideal in itself. He also showed that no finite ring can give an interesting example. The commutative case is also well-known; Kaplansky showed that every finitely generated commutative idempotent ring must have a unit.
Update: Some partial results about this question and a relation to the Wiegold problem in group theory can be found in http://arxiv.org/abs/1112.1802
 A: Consider the ring generated by $a,b,c,d,e$ subject to the relation $a=bc+de$ and all its cyclic shifts: $b=cd+ea$, and 3 more. It is "idempotent" obviously. Can it be killed by one relation? I will ask Agata Smoktunowicz. She should be able to figure it out quickly. 
 Update Agata responded saying that the problem, while interesting, is too difficult. She did try using Groebner-Shirshov bases but without success. She did manage to prove the statement for semigroup algebras using an argument similar to Ozawa's (as Ben Steinberg asked here). If $S$ is a semigroup, $S^2=S$, then $KS$ is generated as an ideal by one element.  
A: What about the ring generated by $a,b,c$ subject to the relations $a^2=a$, $b^2=b$ and $c^2=c$? If this ring is generated as an ideal by a single element $r$ one can show
that $r= \pm a \pm b \pm c + h$ where $h$ is of order at least $2$ (in the ideal generated by 
$ab, ba, ac, \dots$). I can not inagine how this can happen. 
If this works then one can add extra generators which is likely to answer also the second question.
