The following is not an answer to the main question, but provides some context which any answer may need to take into account. This context is certainly well known to many participants in the discussion, but making it explicit may be beneficial.

As the OP says, it is reasonable to conjecture that there is a finitely presented group $G$ and a finitely generated, but infinitely presented, group $H$, that are elementarily equivalent. But confirming this conjecture may be extremely difficult, for the simple reason that it is often extremely difficult to confirm that any pair of finitely generated groups is elementarily equivalent.

The key problem in this area was a famous question of Tarski:

**Question (Tarski, c. 1945):** Is the free group on 2 generators, $F_2$, elementarily equivalent to the free group on 3 generators, $F_3$?

Tarski's question was answered affirmatively by Sela around 2000, but can be seen to be very difficult by at least two different measures. First, it took more than 50 years to answer. Second, Sela's solution spans seven papers and many hundreds, or even thousands, of pages of mathematics. In the subseqent 20 years, the community has been unable to provide any significant simplification of Sela's proof.

So an answer to this question *may* be as difficult as Sela's proof, or even more so. Sela's work extends to all torsion-free hyperbolic groups $\Gamma$, and indeed he is able to classify all finitely generated groups $G$ elementarily equivalent to such $\Gamma$. Unfortunately for this question, he proved that any such $G$ is also hyperbolic, in particular finitely presented. The work of Dahmani, Groves, Guirardel, Hull, Reinfeldt and Weidmann in various combinations begins to extend Sela's techniques to more general classes of "negatively curved" groups, but I think full proofs of elementary equivalence for any of these groups are still a long way off.

In summary, this question may well be extremely difficult, and my *guess* is that it's wide open. If the question is important I would ask Sela himself, and regard his answer as definitive.

"($G$ is abelian) and (every element of $G$ is a square) and (there exists an element of order 2 in $G$)". $\endgroup$13more comments