If the trace of all positive powers of a $n \times n$ complex matrix is $0$, then the matrix must be nilpotent. https://math.stackexchange.com/questions/159167/traces-of-all-positive-powers-of-a-matrix-are-zero-implies-it-is-nilpotent

Is a similar conclusion known to be valid in the setting of finite von Neumann algebras? Let $\tau$ be a faithful normal tracial state on a finite von Neumann algebra $\mathscr{R}$. Let $A \in \mathscr{R}$ be such that $\tau(A^n) = 0$ for $n \in \mathbb{N}$. Does it imply that $A$ is quasinilpotent? Any well-known counter-examples?