It is well-known that the set $\{\sqrt{s}:s\in\mathbb{N}\text{ square-free}\}$ is $\mathbb{Q}$-linearly independent. This implies that, if $f(x)=x^2\in\mathbb{Z}[x]$, then the set of algebraic integers defined by $$f^{-1}[\mathbb{Z}]=\{z\in\mathbb{C}:f(z)\in\mathbb{\mathbb{Z}}\}\subseteq\overline{\mathbb{Z}}$$ contains infinitely many $\mathbb{Q}$-linearly independent numbers. So I was wondering if:
Q: Is it true that $\forall f\in\mathbb{Z}[x]$ monic with $\text{deg}(f)\geq2$ the set $f^{-1}[\mathbb{Z}]\subseteq\overline{\mathbb{Z}}$ contains infinitely many $\mathbb{Q}$-linearly independent algebraic integers? If so, how can one prove it?
Feel free to use as much Galois theory as needed (if needed) or any algebraic number theory apparatus. Also, I guess that the monic condition can be omitted but then we would have to talk about algebraic numbers $\overline{\mathbb{Q}}$ instead.
Update: Sungjin Kim has proven this for $f\in\mathbb{Z}[x]$ with $\text{deg}(f)\geq2$ such that $\exists g\in\mathbb{Z}[x]$ with $\text{deg}(g)\geq1$ and $g(x)^2\ \vert\ f(x)-f(0)$. Since $f$ satisfying the condition of Q implies that $\forall a,b\in\mathbb{Z}:f(x+a)+b$ also satisfies the condition, the previous observation can be generalized for all $f\in\mathbb{Z}[x]$ with $\text{deg}(f)\geq2$ such that $$\exists a\in\mathbb{Z}, g\in\mathbb{Z}[x]\text{ with deg}(g)\geq1:g(x)^2\ \vert\ f(x+a)-f(a)$$ This includes the case where $\exists a\in\mathbb{Z}:f'(a)=0$ as then $x^2\ \vert\ f(x+a)-f(a)$. However, not every integral polynomial of degree $\geq2$ is included here. For example, the polynomial $f(x)=x^2+x$ is such that $\forall a\in\mathbb{Z}:f(x+a)-f(a)=x^2+(2a+1)x$ is always square-free. So how could we include such polynomials? Or is there a counterexample?
Edited (11/1/2023)
If the polynomial $f(z)$ does not have a linear term, then the proof proceeds as follows.
Let $f_0(z)=f(z)-f(0)$. Additionally, assume that $f_0$ does not have a linear term, so $f'(0)=0$. Thus, the proof works for your example $f_0(z)=z^2$, or such as $f_0(z)=z^3+z^2$, $f_0(z)=z^{2023}+11z^2$, etc.
It is enough to show that $$ \mathbb{Q}(f_0^{-1}(p), p \text{ prime}) $$ has infinite degree over $\mathbb{Q}$. Enumerate primes as $p_1<p_2<\ldots$ and consider the sequence of degrees: $$ d_n=[\mathbb{Q}(f_0^{-1}(\{p_1,\ldots, p_n\}):\mathbb{Q}]. $$ If $d_n<d_{n+1}$ for infinitely many $n$, then we are done.
If not, then the sequence $d_n$ eventually stops growing. Let $N<\infty$ be the smallest index that $d_n\leq d_N$ for all $n\in\mathbb{N}$. Consider $D_N=\text{disc}_{\mathbb{Q}} \mathbb{Q}(f_0^{-1}(\{p_1,\ldots, p_N\})$. Then every prime $p$ has a factor $\alpha^2$ with some $\alpha$ in the ring of integers of this field $\mathbb{Q}(f_0^{-1}(\{p_1,\ldots, p_N\})$. By taking $\alpha$ with $f_0(\alpha)=p$ and norm of $\alpha$ in $\mathbb{Q}(\alpha)$ equals $\pm p$, this can be done by taking a root of an irreducible factor of $f_0(z)-p$ whose constant coefficient is $\pm p$. Thus, every prime is ramified in this field. This is a contradiction since ramified primes are those dividing the discriminant $D_N$.
Added proof of a further claim (11/1/2023)
The proof also works for $f_0(z)$ is divisible by $g(z)^2$ with $g(z)\in\mathbb{Z}[z]$ of degree at least $1$.
Let $n$ be the degree of $f_0$, it also is the degree of $f$, and $n\geq 2$. Let $m$ be the degree of $g$, so $2m\leq n$. Let $M$ be the maximum of absolute values of coefficients in $g(z)$ and $f_0(z)$. If we have $$ |z|\leq \frac{p^{1/n}}{(2nM)^{1/n}}, $$ then $|f_0(z)|\leq nM\frac{p}{2nM}=p/2 <p$. Thus, any root $\alpha$ of $f_0(z)=p$ satisfies $$|\alpha|>\frac{p^{1/n}}{(2nM)^{1/n}}.$$ Take sufficiently large prime $p$ so that the right side of above is greater than $1$.
By triangle inequality with isolating leading term of $g$, we have $$|g(\alpha)|\geq |\alpha|^m-mM|\alpha|^{m-1}=|\alpha|^{m-1}(|\alpha|-mM).$$
For prime $p$ large enough, $|g(\alpha)|\geq 2$ is guaranteed, in fact, $p\geq 2nM(2mM)^n$ works. The full set of conjugates of $g(\alpha)$ is a subset of $\{g(\alpha_j), j=1,\ldots, d\}$, where $\alpha_j$ are full set of conjugates of $\alpha$. Since $f_0(\alpha_j)=p$ for each $j=1,\ldots, d$, the norm of $g(\alpha)$ cannot equal $\pm 1$. Thus, $g(\alpha)$ for any $\alpha$ is not a unit in the ring of integers of $K=\mathbb{Q}(f_0^{-1}(\{p_1,\ldots, p_N\})$, provided that $p\geq 2nM(2mM)^n$. Then the square of the principal ideal generated by $g(\alpha)$ is a factor of $p\mathcal{O}_K$, here, $\mathcal{O}_K$ is the ring of integers of $K$. Therefore, any prime $p\geq 2nM(2mM)^n$ is ramified in $K$. This is a contradiction since ramified primes are those dividing the discriminant $D_N$.