= Solution
One implication is immediate: a unitary module equivalence makes each space an isometric copy of the entire other space, which is closed and invariant. For the converse, choose isometric intertwiners $u:H_1\to H_2$ and $v:H_2\to H_1$ supplied by the two embeddings. Their ranges are closed reducing subspaces. We construct the unitary, proving the <Schröder-Bernstein theorem for Hilbert space modules> in this setting.
Put $w=vu$, an isometric intertwiner of $H_1$ into itself, and let $D=H_1\ominus vH_2$. The spaces $w^nD$, $n\geq0$, are mutually orthogonal. For $m>n$, applying the isometry relation reduces their <inner product> to one between $D$ and $w^{m-n}D\subseteq wH_1\subseteq vH_2$, which is zero. Hence
$$
K=\bigoplus_{n\geq0}w^nD
$$
is a closed reducing submodule of $H_1$, and $K=D\oplus wK$. Its <orthogonal complement> lies in $vH_2$, because it is orthogonal to $D$. Combining these decompositions gives
$$
vH_2=wK\oplus K^\perp.
$$
Apply the inverse isometry $v^*$ on $vH_2$ to obtain
$$
H_2=uK\oplus v^*K^\perp.
$$
The operator
$$
\boxed{W\xi=u\xi\quad(\xi\in K),\qquad
W\xi=v^*\xi\quad(\xi\in K^\perp)}
$$
is therefore an isometry onto $H_2$: it is isometric on each orthogonal summand and their images are the two orthogonal summands just exhibited. Both pieces intertwine the action of $M$, as do the <projections> onto $K$ and $K^\perp$, so $W$ is a unitary module equivalence. This proves the reverse implication with no finiteness assumption on either module.
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