Do there exist two elementary equivalent finitely generated groups $G,H$ such that $G$ is finitely presented but $H$ is not finitely presentable?
It seems reasonable to think that finite presentability is not preserved under elementary equivalence among finitely generated groups, so such groups probably exist. However, I do not know any example. The case where $G,H$ have the same universal theory, instead of having the same first order theory, would be already interesting.
The "universal theory" question is easy to solve: indeed if $G$ is a group and $H$ a subgroup, any universal formula true for $G$ is true for $H$. Hence if $G,H$ are groups and both embed into each other, then $G,H$ have the same universal theory.
Denote by $F_2$ the free group on two generators.
Now let $G$ be $F_2\times F_2$ and $H$ the kernel of the homomorphism $G\to\mathbf{Z}$ mapping all four generators to $1$. Then $H$ is not finitely presented, and contains a copy of $G$. So $G,H$ have the same universal theory.
(However they are not EE since $H$ doesn't satisfy the formula expressing: there are elements $x_1,\dots,x_4$ such that each element is uniquely the product of an element in the centralizer of $\{x_1,x_2\}$ and one in the centralizer of $\{x_3,x_4\}$.)
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.
The problem is known and still open but does not seem hopeless. I would start with considering Zilber's old example of two f.g. non-isomorphic nilpotent of class $2$ groups which are e.e.: B.I. Zilʹber, An example of two elementarily equivalent but not isomorphic finitely generated metabelian groups. Algebra i Logika 10 (1971), 309–315. Both of his groups were of course finitely presented. But I would guess that there are similar examples when one of the groups is not finitely presented.
