Concrete cases where $YX=qXY$

The simplest thing to do is just to define and algebra with an element like that:

$R=F\langle X, Y\rangle/(YX-qXY)$ where, say, $q\in F$.

In $M_2(\mathbb R)$, setting $X=Y=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ gives you an example of $XY=qYX$ for any $q\in\mathbb R$ at all.

This is a little generalization of the one you found. This pair works. $$X=\left(\begin{array}{ccc} 1 & a & b\\ 0 & q & q(a d + c - c q)\\ 0 & 0 & q q \end{array}\right),\,Y=\left(\begin{array}{ccc} 0 & 1 & c\\ 0 & 0 & d\\ 0 & 0 & 0 \end{array}\right). $$

A "concrete" algebra where this happens is the "algebra of functions of the quantum plane": let $V$ be the space of functions $\mathbb N_0\to \mathbb C$ and define two elements of $\operatorname{End}_\mathbb C(V)$ as follows: $x$ is the endomorphism that shifts a function, so that $(xf)(n) = f(n+1)$, and $y$ is such that $(yf)(n) = q^n f(n)$. Then $yx=qxy$, and the subalgebra generated by $x,y$ is indeed isomorphic to $\mathbb C\langle x,y\mid yx-qxy\rangle$. If you know some Spanish, check exercise 4.5 here.