Proof that at most one of $e\pi$ and $e+\pi$ can be rational

The proof is correct. For a nit-pick, you should say you have a non-zero polynomial.

Note that it would even suffice to know that at least one of $\pi$ and $e$ is transcendental.

We do not really need to use the fact that both $\pi$ and $e$ are trascendental numbers. If both $e\pi$ and $\pi+e$ would be rational numbers, then $e$ would be a quadratic irrational, so its continued fraction would be eventually periodic due to Lagrange's theorem. However, the continued fraction of $e$ is well-known:

$$ e = [2;1,2,1,1,4,1,1,6,1,1,8,\ldots] $$ and its coefficients are unbounded.