In fact, there is a much stronger result.
Proposition. Let $a,b,c,d\in\mathbb{C}$. If $A,B\in M_n(\mathbb{C})$ satisfy $aA^2+bB^2+cAB+dBA=0_n$, then, generically (for example, randomly choose $a,b,c,d$), $A,B$ are simultaneously triangularizable (for short, ST).
Proof. We follow the last part of the stewbasic's good post. We put $X=\alpha A+\beta B,Y=\gamma A+\delta B$; since generically $\alpha\delta\not= \beta\gamma$, it suffices to show that $X,Y$ are ST.
We obtain (generically) $XY=kYX$ for some complex number $k$. Generically, $k\not=1$ and $k$ is not a primitive root of unity; then, by a result of Drazin, $X,Y$ are ST.
Remark 1. Of course, there are $a,b,c,d$ s.t. $A,B$ are not necessarily ST. For instance, consider $AB+BA=0_2$ with $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\, B=\begin{pmatrix}0&1\\1&0\end{pmatrix}$.
Remark 2. That is linked to the concept of quasi-commutative matrices. There are two non-equivalent definitions:
i) $A,B$ are quasi-commutative iff $AB-BA$ commute with $A,B$. By a result from McCoy, $A,B$ are always ST.
ii) $A,B$ are quasi-commutative iff $AB=kBA$ where $k$ is a complex number. This is the definition we are interested in.