I may be missing something obvious, but is self-adjoint operator in a complex separable Hilbert space (hyper-)maximal?
Lemma: Let $A:D(A) \subsetneq\mathcal H\rightarrow \mathcal H$ self-adjoint, i.e. $A=A^{*}$. Let $B$ an extension of $A$, i.e. $D(A)\subsetneq D(B)$ and $A\phi = B\phi, \forall \phi\in D(A)$. Then $A=B$.
Proof:??
It follows from $\mathscr R(A\pm i)= \mathscr H$, see Theorem 3.79. Corollory 2 here for a proof.