Virtually large groups of small rank (related to 3-manifolds) Edited 25.05.21: the assumptions of the question were incorrect, but as the discussion may be helpful for future MOnauts, I'll strike my mistakes and add clearly marked explanations afterwards.
I am looking for a reason why a 3-manifold group $G$ that is virtually $\mathbb{Z}\times F$, $F$ being either non-cyclic free or a surface group, does not admit a presentation on two generators.
These are the fundamental groups of closed 3-manifolds with $\mathbb{H}^2\times\mathbb{R}$ geometry (Added: thanks @HJRW for pointing out that the strike-through case above corresponds to a non-empty boundary), and it turns out that all other geometries admit examples with fundamental group of rank two, with notable highlight of euclidean geometry where all fundamental groups are virtually $\mathbb{Z}^3$ (and rank two example being the Hantzche–Wendt/Fibonacci manifold). Thus the 3-manifold groups admit examples of virtually high rank groups being nonetheless of small rank themselves. Of course it is well known that a free group on two generators is virtually of arbitrarily high rank.
However, by Boileau & Zieschang, Theorem 1.1, the rank of $\mathbb{H}^2\times\mathbb{R}$ manifolds depends on the genus of the base surface and number of singular fibers of the Seifert fibration (and is at least 3), so apparently being virtually $\mathbb{Z}\times F$ forces the group to be of at least the same rank.
Added: this is my initial confusion - I assumed that the base orbifold of a $\mathbb{H}^2\times\mathbb{R}$ manifold must have genus at least 2, but this is not true. In fact, following the Wikipedia's conventions for Seifert spaces, $\{-1,(o_1,0);(5,1),(5,2),(5,2)\}$ is a $\mathbb{H}^2\times\mathbb{R}$ manifold Seifert-fibering over a shpere, which in particular fits into Theorem 1.1, case ii) of the cited paper (just don't let the initial $g>0$ mislead you) and is indeed of both rank and genus 2. I thank again @HJRW for their comments which got me on the right track eventually. This of course makes the question that followed invalid.
What is the cause that this subgroup bounds the rank of the ambient group from below and, say, free groups or abelian free $\mathbb{Z}^3$ do not? I would be happy if there is a geometric 3-dimensional reason in play here, but would be grateful for refreshing my general group theory as well.
 A: The question stems from a misinterpretation of Theorem 1.1 in the paper by Boileau  and Zieschang. Theorem 1.1 excludes a fair number of cases, in particular, it does not apply to (totally oriented) closed Seifert manifolds with 3 singular fibers and base of genus 0. Some of these excluded Seifert manifolds provide counter-examples to your claim about rank $\ge 3$.
For instance, take the exterior $N$ of a $(p,q)$-torus knot which is nontrivial and not the trefoil. The genus of this knot is
$$
g=\frac{(p-1)(q-1)}{2}\ge 2
$$
(because I excluded the trefoil which has genus 1). The manifold $N$ is a surface bundle over the circle whose fiber $F$ is the once-punctured surface of genus $g$. The monodromy of this fibration is a finite order (actually, the order is $pq$) homeomorphism $h: F\to F$. Thus, if we collapse the boundary of $F$ to point, we obtain a closed surface $S$ of genus $g$ and $h$ will project to a finite order homeomorphism $f: S\to S$. The mapping torus $M=M_f$ is a Seifert manifold of type ${\mathbb H}^2\times {\mathbb R}$ obtained by a Dehn filling of the boundary of $N$. The base of the Seifert fibration will have three singular points and genus 0: Two of the singular fibers come from $N$ and one comes from the solid torus attached to $\partial N$ as the result of our Dehn filling.
(It is a general fact that the mapping torus of a finite order homeomorphism of a hyperbolic surface is a Seifert manifold of type ${\mathbb H}^2\times {\mathbb R}$.) Since the group $\pi_1(N)$ is 2-generated, the quotient group $\pi_1(M)$ is also 2-generated.
