$e$ 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?
We do not really need to use the fact that both $\pi$ and $e$ are trascendental numbers. If both $e\pi$ and $\pi+e$ were 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.