No. Let

Here's a quick test which might disprove your hopes very quickly:

Take $y$ 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 entries determinant $d_i$. Then 1$, for example, $$[x,y,y]_{ij} = (d_i-d_j)^2 x_{ij}.$$ In particular\mathrm{diag}(17, 1/17)$. Write out your relation, let leaving all the elements of $y = \mathrm{diag}(1,2,3,4,\ldots,n)$. Choose y$ as variables. After clearing denominators, you have $x_{ij}$ to be zero if n^2$ simultaneuous homogenous equations in $|i-j| \neq 1$n^2$ variables. (If I haven't made any dumb errors, and they have degree $3n$.) Ask your favorite computer algebra system to be arbitrary otherwisesolve them for you. Then $[x,y,y]=x$. For generic values If any of the $x_{i \ i \pm 1}$, roots are not on the matrix hypersurface $x$ is invertible but not unipotent.\det y=0$, then you have a counterexample!

No. Let $y$ be the diagonal matrix with entries $d_i$. Then $$[x,y,y]_{ij} = (d_i-d_j)^2 x_{ij}.$$ In particular, let $y = \mathrm{diag}(1,2,3,4,\ldots,n)$. Choose $x_{ij}$ to be zero if $|i-j| \neq 1$, and to be arbitrary otherwise. Then $[x,y,y]=x$. For generic values of the $x_{i \ i \pm 1}$, the matrix $x$ is invertible but not unipotent.