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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 (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 examples being the Fibonacci manifolds). 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 being virtually $\mathbb{Z}\times F$ forces the group to be of at least the same rank.

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.

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.

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, 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 examples being the Fibonacci manifolds). 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 being virtually $\mathbb{Z}\times F$ forces the group to be of at least the same rank.

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.

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 (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 examples being the Fibonacci manifolds). 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 being virtually $\mathbb{Z}\times F$ forces the group to be of at least the same rank.

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.