Traces of powers of a matrix $A$ over an algebra are zero implies $A$ nilpotent.

I would like to have a result similar to "Traces of all positive powers of a matrix are zero implies it is nilpotent". Namely:

Let $R$ be a commutative $\mathbb{C}$-algebra, $A \in \mathcal{M}_n(R)$ such that $\mathrm{tr}(A)=\cdots=\mathrm{tr}(A^n)=0$. Does it follow that $A^n=0$?

I have no idea weather or not the analogy of eigenvalue exists for matrices over algebra so I can't simply generalize considerations from cited post. Any help will be appreciated.

The answer to your question is positive. For a self-contained proof, see Corollary 4.1 (b) in my note The trace Cayley-Hamilton theorem.