Here's a strategy that I've been toying with. It seems unlikely to me that it works, but perhaps it'll inspire someone to have a better idea. Fix an embedding of $\overline{\mathbb{Q}}$ into $\mathbb{C}$, thereby fixing Try to arrange a complex conjugation $c$. Let $L$ be the fixed field of $\overline{\mathbb{Q}}$ under a Sylow $3$-subgroup of the absolute Galois group of $\mathbb{Q}$, and let $K$ be the fixed field of $L$ under $c$. Then the whose Galois group of $K$ is an extension of has trivial pro-$5$ Sylow subgroup but nontrivial pro-$p$ Sylow subgroup for $\mathbb{Z}/2\mathbb{Z}$ by a big pro-$3$ group. p=2,3$. Let $f$ be the product of an irreducible quadratic and an irreducible cubic over $K$. Since there are no irreducible quintics over $K$, for each $a \in K$, the polynomial $f-a$ either has a root, or factors as an irreducible quadratic times an irreducible cubic. We'd be done if we can arrange for the latter to occur only finitely many times. But this seems like a stretch....

Here's a strategy that I've been toying with. It seems unlikely to me that it works, but perhaps it'll inspire someone to have a better idea. Fix an embedding of $\overline{\mathbb{Q}}$ into $\mathbb{C}$, thereby fixing a complex conjugation $c$. Let $L$ be the fixed field of $\overline{\mathbb{Q}}$ under a Sylow $3$-subgroup of the absolute Galois group of $\mathbb{Q}$, and let $K$ be the fixed field of $L$ under $c$. Then the Galois group of $K$ is an extension of $\mathbb{Z}/2\mathbb{Z}$ by a big pro-$3$ group. Let $f$ be the product of an irreducible quadratic and an irreducible cubic over $K$. Since there are no irreducible quintics over $K$, for each $a \in K$, the polynomial $f-a$ either has a root, or factors as an irreducible quadratic times an irreducible cubic. We'd be done if we can arrange for the latter to occur only finitely many times. But this seems like a stretch....