Invertible matrices satisfying $[x,y,y]=x$ I have been thinking about this question for quite some time but now  this  question by Denis Serre revived some hope. 
 Question.  Let $x,y$ be invertible matrices (say, over $\mathbb C$) and $[x,y,y]=x$ where $[a,b]=a^{-1}b^{-1}ab$, $[a,b,c]=[[a,b],c]$. Does it follow that some power of $x$ is unipotent? 
The motivation is this. Consider the one-relator group $\langle x,y \mid [x,y,y]=x\rangle$. It is hyperbolic (proved by A. Minasyan) and residually finite (that is proved in my paper with A. Borisov). If the answer to the above question is "yes", then that group would be non-linear which would provide an explicit example of non-linear hyperbolic group. 
 Update 1.  Can $x$ in the above be a diagonal matrix and not a root of 1?
 Update 2.  The group is residually finite, so it has many representations by matrices such that $x, y$ have finite orders (hence their powers are unipotents). 
 Update 3.  The group has presentation as an ascending HNN extension of the free group: $\langle a,b,t \mid a^t=ab, b^t=ba\rangle$. So it is related to the Morse-Thue map. Properties of that map may have something to do with the question. See two quasi-motivations of the question as my comments .
 A: Here's a quick test which might disprove your hopes very quickly:
Take $n$ to be small: Try $2$ first, and $5$ is probably near the limit of a computer algebra system. Choose $x$ to be a random $n \times n$ diagonal matrix with determinant $1$, for example, $\mathrm{diag}(17, 1/17)$. Write out your relation, leaving all the elements of $y$ as variables. After clearing denominators, you have $n^2$ simultaneuous homogenous equations in $n^2$ variables. (If I haven't made any dumb errors, they have degree $3n$.) Ask your favorite computer algebra system to solve them for you. If any of the roots are not on the hypersurface $\det y=0$, then you have a counterexample!
A: The answer is "No". Indeed, consider the 1-related group $G=\langle x,y \mid [x,y,y]=x\rangle$ That group has a presentation $\langle a,b,t \mid a^t=ab, b^t=ba\rangle$ (easy to check). Thus it is an ascending HNN extension of the free group. The  group $G$ is hyperbolic (proved by Minasyan using the Bestvina-Feighn combination theorem). By a theorem of Hagen and Wise, the group (and almost every other hyperbolic ascending HNN extensions of a free group) acts geometrically on a finite dimensional CAT(0)-cube complex. By a result of Ian Agol then the group $G$ is virtually special and hence linear, i.e., there exists an injective homomorphism $\phi$ from $G$ to a special linear group (over $\mathbb{Z}$). Since $G$ is hyperbolic, it does not contain nilpotent non-Abelian subgroups. Thus  the pair of integer matrices  $(\phi(a), \phi(b))$ is an example showing that the answer is "no". 
