= Solution
Over $\mathbb C$, triangularize $B$ and denote its eigenvalues, with algebraic multiplicity, by $\beta_1,\ldots,\beta_n$. Then
$$
\chi_{B^m}(z^m)=\prod_{j=1}^n(\beta_j^m-z^m).
$$
The numbers $\omega_l=e^{2\pi il/m}$ are all the <root of unity> solutions of $w^m=1$, so
$$
\prod_{l=1}^m(t-\omega_lz)=t^m-z^m.
$$
Consequently
$$
\begin{aligned}
\prod_{l=1}^m\chi_B(\omega_lz)
&=\prod_{j=1}^n\prod_{l=1}^m(\beta_j-\omega_lz)\\
&=\prod_{j=1}^n(\beta_j^m-z^m)
=\boxed{\chi_{B^m}(z^m)}.
\end{aligned}
$$
If $\lambda$ is an eigenvalue of $B^m$, choose $z$ with $z^m=\lambda$. The displayed identity makes at least one factor $\chi_B(\omega_lz)$ zero. Thus $\mu=\omega_lz$ is an eigenvalue of $B$ and
$$
\boxed{\mu^m=\lambda}.
$$
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