Polynomial bijections from $\mathbb{Q}$ to $\mathbb{Q}$

Here’s yet another effort:

We may assume that our polynomial $f(x)\in\Bbb Q[x]$ is monic, just by dividing the whole polynomial by the coefficient of the highest-degree term $x^d$.

Now let $p$ be a prime not dividing any of the numerators of the coefficients of $f$. I claim that there is no $z\in\Bbb Q$ with $f(z)=1/p$. It’s the proof of Eisenstein upside-down:

Let $z=m/n$, with coprime integers $m$ and $n$. If $p\nmid n$, then $p\nmid f(z)$, while if $p\mid n$, then $p^d$ is exactly the power of $p$ dividing the numerator of $f(z)$. (In $p$-adic language, if $v_p(z)\ge0$, then $v_p(f(z))\ge0$, while if $v_p(z)<0$, then $v_p(f(z))=dv_p(z)$.)

In any case, the numerator of $f(z)$ can not be only singly divisible by $q$.

Assume $f(x)=a_nx^n+\ldots +a_1x+a_0\in\Bbb Q[x]$ with $a_n\ne0$ induces a surjection $\Bbb Q\to \Bbb Q$. Let $p$ be a large prime such that $|a_i|_p=1$ for all non-zero $a_i$. Then $$|f(x)|_p\le\max\{\,|a_kx^k|_p\mid k\ge0\,\}=\max\{\,|x|^k_p\mid a_k\ne 0\,\}$$ with equality if these are distinct, i.e., if $|x|_p\ne 1$. In particular, either $|x|_p\le 1$ and so $|f(x)|_p\le 1$, or $|x|_p\ge p$ and so $|f(x)|_p\ge p^n$. For surjectivity, we need some $x\in\Bbb Q$ with $|f(x)|_p=p$. Therefore we need $n\le 1$.

You could use the Hilbert irreducibility theorem: let $p(X) \in \mathbb Q[X]$ be such a polynomial, and consider the polynomial

$$f(Q,X) = p(X) + Q\in\mathbb Q(Q)[X].$$

This is irreducible over $\mathbb Q(Q)$: by Gauss' lemma it is irreducible as soon as it is irreducible in $\mathbb Q[Q,X] = \mathbb Q[X][Q]$, and as a polynomial in $Q$ it is monic and linear.

As the conditions for the Hilbert irreducibility theorem are met, it can be concluded that there exists a $q\in\mathbb Q$ such that $f(q,X)$ is irreducible. In particular $p(X) + q$ does not have a rational root, so $-q$ is not in the image.