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.

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!