Here is a complement to YCor's answer, adding some references.

**Notation.** For $G$ a group and $x$, $y$ two elements of $G$, we set $x^y := y^{-1}x y$ and $[x, y] := x^{-1}y^{-1}xy$.

**Definition.** A group is said to have *finite Prüfer rank $r$* if every finitely generated subgroup can be generated by $r$ elements and $r$ is the least such integer.

The answer to the question stated in the body of the post is **no**, as witnessed by the following **finitely presented** counter-example due to Baumslag and Remeslennikov [1, Theorem 11.1.5]:

**Claim.** Let $$G = \langle a, s, t \, \vert \, a^t = a a^s, [a, a^s] = 1 = [s,t] \rangle$$ be the Baumslag–Remeslennikov group. Then $G$ is a finitely presented torsion-free metabelian group with subgroup $\langle a, s \rangle \simeq \mathbb{Z} \wr \mathbb{Z}$.

*Sketch of the claim's proof.* Check that $[G, G] = \langle a^{s^i} \,\vert\, i \in \mathbb{Z} \rangle \simeq \mathbb{Z}^{(\mathbb{Z})}$. It follows immediately that $G$ is torsion-free and metabelian. See [1, Proof of Theorem 11.1.5] for details.

**Note.** More generally, a theorem of Baumslag and Remeslennikov (1973) states that every finitely generated metabelian group can be embedded into a finitely presented metabelian group. Even more to the point is Thomson's theorem (1977) [2]: every finitely generated linear soluble group can be embedded into a finitely presented linear soluble group. Starting from $H = \mathbb{Z} \wr \mathbb{Z}$, Thomson's theorem yields a linear and finitely presented soluble group $G$ containing $H$. This group is made explicit in the above claim.

**Corollary.** The Baumslag–Remeslennikov group, i.e., the group $G$ of the above claim, is $\mathbb{C}$-linear but not $\mathbb{Q}$-linear.

The proof of the previous corollary relies on the following two lemmas.

**Lemma 1. (YCor's key argument)** A finitely generated soluble $\mathbb{Q}$-linear group has finite Prüfer rank.

**Lemma 2.** (Levic–Remeslennikov, 1969 [3] and [4])
A finitely generated torsion-free metabelian group is $\mathbb{C}$-linear.

*Proof of Lemma 1.* Because of Mal'cev's Theorem (1951), see e.g. [1, Theorem 3.1.6.ii], it suffices to show that any finitely generated subgroup of $T_n(\overline{\mathbb{Q}})$ ($n \ge 1$), the group of invertible upper $n \times n$ triangular matrices over the algebraic closure of $\mathbb{Q}$, has finite Prüfer rank. Such a subgroup $H$ is a finitely generated subgroup of $T_n(K)$ for some number field $K = K(H)$. It is therefore the extension of a subgroup $N$ of $U_n(K)$, the $n \times n$ unipotent matrices over $K$ by a finitely generated subgroup $Q$ of $(K^{\times})^n = (K \setminus \{0\})^n$. Since $Q$ is a finitely generated Abelian group, it has finite Prüfer rank and so has $N$ since the additive group of $K$ is a finite dimensional $\mathbb{Q}$-linear space. Now, it is an easy exercise to show that the property of having finite Prüfer rank is stable under taking extensions.

*Proof of the corollary.* The Baumslag–Remeslennikov group is $\mathbb{C}$-linear by the Claim and Lemma 2. It cannot be $\mathbb{Q}$-linear because of Lemma 1. Indeed, it contains a subgroup of infinite Prüfer rank, namely $\langle a, s \rangle \simeq \mathbb{Z} \wr \mathbb{Z} \supset \mathbb{Z}^{(\mathbb{Z})}$.

**Note.** As pointed out by YCor in a comment, a simple and direct way to show that the Baumslag-Remeslennikov group $G$ is linear is to leverage the semi-direct decomposition $$G \simeq \mathbb{Z}^{(\mathbb{Z})} \rtimes \mathbb{Z}^2 \subset \mathbb{Z}[s^{\pm 1}] \rtimes (\mathbb{Z}[s^{\pm 1}, t^{\pm 1}])^{\times}.$$ The Magnus Embedding Theorem (1939), see e.g. [1, 11.3.2] may serve a similar purpose in more general situations.

- [1] J. Lennox and D. Robinson, "The Theory of Infinite Soluble Groups", 2004.
- [2] M. Thomson, "Subgroups of Finitely Presented Solvable Linear Groups", 1977.
- [3] E. Levic, "Representation of soluble groups by matrices over a certain field of characteristic zero", 1969.
- [4] V. Remeslennikov, "Representation of finitely generated metabelian groups by matrices", 1969.