I'll show how to prove the more general case with complex matrices: find the maximum of $\operatorname{Tr}(UZ)$ over all unitaries $U$:
$$\max_{U: U^\dagger U=I}|\operatorname{Tr}(UZ)|.$$
Leveraging the polar decomposition, we know that $Z$ can be always written as $Z=VP$ for some unitary $V$ and positive semi-definite $P\ge0$.
Because the product of unitaries is another unitary, this observation reduces the problem to the following:
$$\max_{U: U^\dagger U=I}\operatorname{Tr}(UP).$$
Moreover, observe that the $P$ in the polar decomposition equals $\sqrt{A^\dagger A}$. Denoting with $\{u_k\}$ the eigenvectors of $P$, and $s_k$ the eigenvalues of $P$ (i.e. the singular values of $Z$), we have
$P=\sum_k s_k u_k u_k^*,$
and therefore for every unitary $U$ there is an orthonormal basis $\{v_k\}$ such that
$$UP=\sum_k s_k v_k u_k^*.$$
It follows that the trace reads
$\operatorname{Tr}(UP)=\sum_k s_k \langle u_k,v_k\rangle$, and taking the absolute value,
$$
\lvert\operatorname{Tr}(UP)\rvert=\left\lvert\sum_k s_k
= |\langle u_k,v_k\rangle|\right\rvert
\le \sum_k s_k \lvert\langle u_k,v_k\rangle\rvert
\le\sum_k s_k
= \operatorname{Tr}P.
\tag X
$$
therefore if there is a $U$ such that $\lvert\operatorname{Tr}(UP)\rvert=\sum_k s_k$, that ought to be maximum we are looking for.
But finding this $U$ is trivial at this point: just use $U=V^\dagger$.