$e$ is not a quadratic irrational

Assume. $ae^2+be+c=0$ with $a,b,c\in\mathbb{Z}$, not all zero.

Multiply by $q!$. For suitable $q$, $q!(ae^2+be+c)$ is an integer.

Series. Expand $e^2=\sum 2^n/n!$ and $e=\sum 1/n!$. The sum $a\sum 2^nq!/n!+b\sum q!/n!+c$ is an integer.

Tail bound. The tail involves $\sum_{n>q}\frac{(a2^n+b)q!}{n!}$, which is bounded by a constant times $2^{q+1}/(q+1)$ — but $c$ is a fixed integer, so for large $q$, the total is strictly between $0$ and $1$, contradicting integrality.

Conclusion. $e$ is not quadratic irrational. (In fact, $e$ is transcendental.)