Can't prove the following:
Let $k$ be a field, $A$ an associative (non unital) $k$-algebra in which every element is nilpotent, and $A$ is finitely dimensional over $k$. Then for every commutative unital $k$-algebra $R$ every element of $A \otimes_k R$ is nilpotent.
If one uses the Wedderburn Theorem as it is stated here (forget about the characteristic $0$!), then your question is almost solved: the $m$th power of a sum $\sum_{i=1}^na_i\otimes r_i$ is a sum of elements of the form $a_{i_1}\cdots a_{i_m}\otimes r_{i_1}\cdots r_{i_m}$, and since $A$ is nilpotent of finite index, say $m$, then all products $a_{i_1}\cdots a_{i_m}$ are $0$.
