$e$ and $\pi$ are rather peculiar numbers. It turns out that, in addition to being irrational numbers, they are also transcendental numbers. Basically, a number is transcendental if there are no polynomials with rational coefficients that have that number as a root.
Clearly, $p(x) = (x-e)(x-\pi)$ is a polynomial whose roots are $e$ and $\pi$, so its coefficients cannot all be rational, by the definition of transcendental numbers. Expanding that expression, we get
$$(x-e)(x-\pi) = x^2 - (e+ \pi)x + e\pi$$
This means that $1, -(e+\pi), e\pi$ cannot all be rational. If all the coefficients were rational, we would have found a polynomial with rational coefficients that had $e$ and $\pi$ as roots, and that has been proven impossible already. Hermite proved that $e$ is transcendental in 1873, and Lindemann proved that $\pi$ is transcendental in 1882. In fact, Lindemann's proof was similar to Hermite's proof and was based on the fact that $e$ is also transcendental.
In other words, at most one of $e+\pi$ and $e\pi$ is rational. (We know that they cannot both be rational, so that's the most we can say). Are there any more conditions required for this proof to be correct?