Shortlisted Problems for the Romanian NMO

Let $z=(yx{)}^{n-1}$. Then

$xyxz=xy(yx{)}^{n-1}z=x(yx{)}^{n}z=x\cdot 0\cdot z=0$.

Now, suppose $xz=0$. Then $x(yx{)}^{n-1}=0$. But $xyx\ne 0$, and $(xy{)}^n=0$ implies $(xy{)}^{n-1}xy=0$. If $x(yx{)}^{n-1}=0$, then $xyx=0$ (since $n\ge 2$), which contradicts the hypothesis. Therefore, $xz\ne 0$.

Thus, such $z$ exists.