**Second try** deals with a very particular case of the problem for $n=2$ but, at least, it includes the baby example (solved in the other answer). First, I would like to slightly reformulate the question.

**Problem.** Let $f:{\mathbb A}_{\mathbb C}^1\to{\mathbb A}_{\mathbb C}^n$ be a polynomial map, let $I\subset A:={\mathbb C}[x_1,\dots,x_n]$ be the ideal of all polynomials vanishing on the image of $f$, and let $V\subset A$ be a
finite-dimensional ${\mathbb C}$-linear subspace. Denote
$A_0:={\mathbb Q}[x_1,\dots,x_n]$. One has to prove that
$\dim_{\mathbb Q}(I+V)\cap A_0=\infty$ implies $\dim_{\mathbb Q}(I+{\mathbb C})\cap A_0>1$
(and, therefore, $\dim_{\mathbb Q}(I+{\mathbb C})\cap A_0=\infty$).

This is quite an intriguing problem, in particular, because there is a feeling that there are no developed tools to solve it. (Though, I think that there is a counter-example.)

When $n=2$, the ideal $I$ is generated by an irreducible polynomial $g$ (in the baby example, $g=x_2-\sqrt2x_1-r$ with a transcendent $r$) whose leading homogeneous part has the form
$g_0:=x_1^{m-k}\displaystyle\prod_{i=1}^k(x_2-c_ix_1)$; we say that the $c_i$'s and $\infty$ (the latter, with multiplicity $m-k$) are roots of $g_0$. If one of the $c_i$'s is not algebraic over ${\mathbb Q}$, we are done because the leading homogeneous part of $gh$ can belong to $A_0$ only if $h=0$. On the other hand, if all coefficients of $g$ are algebraic over ${\mathbb Q}$, then
$\overline x_1,\overline x_2\in\overline{\mathbb Q}[x_1,x_2]/I$ are algebraically dependent over $\overline{\mathbb Q}$ and, hence, over
${\mathbb Q}$, implying $I\cap A_0\ne0$.

Here, we consider a first nontrivial particular case of $g=g_0-r$ with transcendent $r\in{\mathbb C}$. Suppose that
$\displaystyle\sum_{i=0}^{m+l}q_i=(g_0-r)\sum_{i=0}^la_i$, where $a_i,q_i$ are homogeneous of degree $i$, $a_l,q_{m+l}\ne0$, and $q_i\in A_0$ for all $i\ge s$.
It suffices to show that $(m+1)(s-1)\ge l$.

We have
$$g_0a_l=q_{m+l},\qquad q_{l-im}=g_0a_{l-(i+1)m}-ra_{l-im},\quad0\le i\le\frac{l-s}m.$$

Since $g_0a_l=q_{m+l}\in A_0$, the roots of $a_l$ and, therefore, the coefficients of $a_l$ are algebraic ($\infty$ is considered as algebraic). Let us show that $g_0$ divides $a_l$. Indeed, denote by $d:=\text{gcd}(g_0,a_l)$ the greatest common divisor of $g_0$ and $a_l$, and let $c$ be a root of $\frac{g_0}d$. Substituting $x_2:=cx_1$ (if $c=\infty$, we substitute $x_1:=0$) in $\frac{q_l}d=\frac{g_0}da_{l-m}-r\frac{a_l}d$, we obtain $\frac{q_l}d|_{x_2:=cx_1}=-r\frac{a_l}d|_{x_2:=cx_1}$, where the polynomials $\frac{a_l}d|_{x_2:=cx_1}\ne0$ and $\frac{q_l}d|_{x_2:=cx_1}$ have algebraic coefficients. This contradicts the assumption that $r$ is transcendent. Consequently, $a_{l-m}=\frac{q_l}{g_0}+r\frac{q_{m+l}}{g_0^2}$.

By induction, suppose that
$$a_{l-im}=\sum_{j=0}^ir^j\frac{q_{l+(j-i+1)m}}{g_0^{j+1}}$$
with $q_{l+(j-i+1)m}$ divisible by $g_0^{j+1}$ for all $0\le j\le i$, where $0\le i\le\frac{l-s}m$. From $q_{l-im}=g_0a_{l-(i+1)m}-ra_{l-im}$, we obtain
$q_{l-im}=g_0a_{l-(i+1)m}-\displaystyle\sum_{j=0}^ir^{j+1}\textstyle\frac{q_{l+(j-i+1)m}}{g_0^{j+1}}$.
If one of the polynomials $\frac{q_{l+(j-i+1)m}}{g_0^{j+1}}$, $0\le j\le i$, is not divisible by $g_0$, we divide the equality by the greatest common divisor $d$ of $g_0$ and these polynomials. As above, we arrive at a contradiction after substituting a root of $\frac{g_0}d$. Consequently,
$a_{l-(i+1)m}=\frac{q_{l-im}}{g_0}+\displaystyle\sum_{j=0}^ir^{j+1}\textstyle\frac{q_{l+(j-i+1)m}}{g_0^{j+2}}$.

Thus, we conclude that $q_{l+m}$ is divisible by $g_0^{i_0+2}$, where
$i_0:=[\frac{l-s}m]$.

There exists a homogeneous polynomial $g_1\in A$ such that $g_0g_1\in A_0$ and $g_0h\in A_0$ for a homogeneous $h\in A$ implies $h\in g_1A_0$. Such a minimal $g_1$ can be easily constructed in the form
$g_1=\displaystyle\prod_{j=1}^t(x_2-c'_jx_1)$, where $c'_j$ are conjugated to suitable $c_i$'s so that the total collection of conjugated $c$'s in $g_0g_1$ is "complete" thus representing $g_0g_1$ as a product of polynomials of the form $x_1^{\deg p}p(\frac{x_2}{x_1})$, where $p$ is the minimal polynomial of some $c_i$. Note that $\deg g_1\ge1$ as, otherwise, $g_0\in A_0$ and
${\mathbb Q}+{\mathbb Q}g_0\subset(I+{\mathbb C})\cap A_0$.

It follows that $q_{l+m}$ is divisible by $(g_0g_1)^{i_0+2}$, so
$l+m\ge(m+1)(i_0+2)>(m+1)(\frac{l-s}m+1)$ and $(m+1)(s-1)\ge l$.

**First try (stupid and wrong).** Let $f_1(t):=t+\pi$ and $f_2(t):=t$. As $f_1$ and $f_2$ are algebraically independent over ${\mathbb Q}$, the abelian group $M_0$ consists only of constants. On the other hand, the polynomials $h_m(x_1,x_2):=(x_1-x_2)^mx_2$, $m\in{\mathbb N}$, all belong to $M_1$. So, $M_1$ is not a finitely generated abelian group.