Suppose such a polynomial exists. Consider it as a morphism $f:\mathbb A^1_k \to \mathbb A^1_k$. You can compactify it as $g:\mathbb P^1_k\to \mathbb P^1_k$. But then the image of $g$ is a closed subscheme of $\mathbb P^1_k$ so it is either $\mathbb P^1_k$ or a finite scheme $S$ over $k$. You have a contradiction.

I'm probably missing though

Suppose such a polynomial exists. Consider it as a morphism $f:\mathbb A^1_k \to \mathbb A^1_k$. You can compactify it as $g:\mathbb P^1_k\to \mathbb P^1_k$ by setting $g(\infty) = \infty$. This is a proper morphism. Its image is either $\mathbb P^1_k$ or a finite scheme $S$ over $k$. You have a contradiction.

Am I missing something?