Given
$$A_1^k + A_2^k + \cdots + A_m^k = 0, \qquad \forall k \in \mathbb N^+$$
then
$$\mbox{Tr}(A_1^k) + \mbox{Tr}(A_2^k) + \cdots + \mbox{Tr}(A_m^k) = 0$$
where $A_1, A_2, \dots, A_m$ are $n\times n $ matrices. Would that mean that all the eigenvalues of these matrices are $0$?
The book where this is stated says it could be proved by using Newton's formulae but I haven't heard of those before.
Write the equalities linking the traces for $k=1,\cdots ,mn$ and define the $(\lambda_i)_{i\leq mn}$ as follows
$spectrum(A_1)=(\lambda_1,\cdots ,\lambda_n),spectrum(A_2)=(\lambda_{n+1},\cdots,\lambda_{2n})$ and so on.
Use the fact that $tr(A_1^k)=\sum_{i\leq n}\lambda_i^k$. We obtain a system of $mn$ equations with $mn$ complex unknowns. The sole solution is the trivial one and we are done. This can be proved using the Newton's identities. cf on this website
"Traces of all positive powers of a matrix are 0 implies it is nilpotent".
Remark. The above result is still valid over a field of characteristic $> mn$.
A question to the OP. In your book, does the author study the possible simultaneous triangularization of the $(A_i)$ ?