Are $2^n\pm1$ and $n!\pm1$ powerful for only finitely many $n$? (Erdős #936)

An integer $m$ is powerful if $p\mid m\Rightarrow p^2\mid m$ for every prime $p$. Is each of $2^n\pm 1$ and $n!\pm 1$ powerful for only finitely many $n$? That is, are there only finitely many $n$ for which $2^n-1$ is powerful, only finitely many for which $2^n+1$ is powerful, and likewise only finitely many for which $n!-1$ is powerful and only finitely many for which $n!+1$ is powerful?

Acceptance. FULLY RESOLVES: an unconditional proof — machine-checkable preferred, else a full written proof — that each of $2^n\pm1$ and $n!\pm1$ is powerful for only finitely many $n$; OR a disproof exhibiting an infinite family of $n$ for which one of these is powerful, with a proof that the family works. ADVANCES: an unconditional finiteness proof for any single one of the four sequences ($2^n-1$, $2^n+1$, $n!-1$, $n!+1$); a newly discovered $n$ making one of them powerful (with prime factorisation certifying the powerful condition) beyond the currently tabulated examples; an extension of the exhaustively verified range with reproducible code and a certificate of completeness; or a sharper abc-conditional statement (e.g. an effective bound on the largest such $n$). Deliver the proof, or the certified witness, or the search code together with the attained bound.