Algebraic independence of $e$ and $\pi$

Individual transcendence. $e$ (Hermite) and $\pi$ (Lindemann) are both transcendental.

Algebraic independence. Unknown whether $e$ and $\pi$ are algebraically independent over $\mathbb{Q}$. This would require showing no nonzero polynomial $P\in\mathbb{Q}[X,Y]$ satisfies $P(e,\pi)=0$.

Known. At most one of $e+\pi$ and $e\pi$ can be rational (since if both were, $e,\pi$ would be algebraic of degree $\le2$). But even this is unknown.

Conclusion. Algebraic independence of $e,\pi$ is a major open problem.