Pattern of divisibility within $2^n+1\equiv0\pmod{n}$
Not an answer, just a comment so far.
Maybe the patterns in this tabulated view give additional insight.
n o(n) factors | factors
of n | of w(n)=(2^n+1)/n
------------------------|------------------------------------------------------
[ 3, 2, "3", "3"]
[ 9, 6, "3^2", "3.19"]
[ 27, 18, "3^3", "3.19. 87211"]
[ 81, 54, "3^4", "3.19.163. 87211.135433. 272010961"]
[ 243, 162, "3^5", "3.19.163. 1459. 87211.135433.139483. <big>"]
[ 729, 486, "3^6", "3.19.163. 1459. 87211.135433.139483. <big>"]
[ 2187, 1458, "3^7", "3.19.163. 1459.17497. 87211.135433.139483. <big>"]
[ 6561, 4374, "3^8", "3.19.163. 1459.17497.52489.87211.135433.139483. <big>"]
... ... ... ...
[ 171, 18, "3^2.19", "3.19. 571. 174763. <big>"]
[ 513, 18, "3^3.19", "3.19. 571. 87211. 144667.174763. <big>"]
[ 1539, 54, "3^4.19", "3.19.163. 571. 87211.135433. 144667.174763. <big>"]
[ 4617, 162, "3^5.19", "3.19.163. 571. 1459. 87211.135433.139483.144667.174763. <big>"]
[13851, 486, "3^6.19", "3.19.163. 571. 1459. 87211.135433.139483.144667.174763. <big>"]
... ... ... ...
[ 3249, 342, "3^2.19^2", "3.19. 571. 174763. <big>"]
[ 9747, 342, "3^3.19^2", "3.19. 571. 87211. 144667.174763. <big>"]
... ... ... ...
[13203, 162, "3^4.163", "3.19.163. 8803. 87211.135433. 196579.<big>"]
... ... ... ...
Here "o(n)" means the group-order such that $n |2^k-1 $ where $o(n):=k \text{ using the minimal } k $ (for instance znorder(Mod(2,n)) in Pari/GP).
Note that $w(n) = { 2^n+1\over n}$ is also $w(n) = { 2^{2n}-1\over (2^{n}-1) n}$ which might be easier for the description of the involved primefactors: "all that which occur in $2^{2n}-1$ but not in $2^n-1$ and also use only cases where $n|2^{2n}-1$"
I'd also suggest to check, whether there are cases where $w(n)$ is not squarefree. Maybe there is something there like with the Wieferich-primes.
Using the idea of the proof of Proposition 3, we have the following :
If $n$ is of the form $mN^k$ where $k$ is a positive integer, and $N$ is a prime, and $m\ (\gt 1)$ is an integer satisfying $\gcd(m,N)=1$, then $n$ is divisible by a prime $p$ such that $p|2^{N^{k}\gcd(m,\frac{N-1}{2})}+1$.
The proof is written at the end of the answer.
Letting $N=19$, we have the following :
If $n$ is of the form $m\cdot 19^k$ where $k$ is a positive integer, and $m\ (\gt 1)$ is a positive integer satisfying $\gcd(m,19)=1$, then $n$ is divisible by a prime $p$ such that $p|2^{9\cdot 19^{k}}+1$.
(Since the paper shows that if $n\gt 3$, then $9|n$, we have $\gcd(m,\frac{19-1}{2})=9$.)
The prime factorization of $2^{9\cdot 19^1}+1$ is given by $$2^{9\cdot 19^1}+1$$ $$=3^3×19^2×571×174763×160465489×19177458387940268116349766612211$$
So, we can say that if $171|n$ and $n\not=171$, then $n$ is divisible by at least one of $$513\ (=3^3\times 19)$$ $$3249\ (=3^2\times 19^2)$$ $$97641\ (=3^2\times 19\times 571)$$ $$29884473\ (=3^2\times 19\times 174763)$$ $$27439598619\ (=3^2\times 19\times 160465489)$$ $$3279345384337785847895810090688081\ (=3^2\times 19\times 19177458387940268116349766612211)$$
Finally, let us prove the following :
If $n$ is of the form $mN^k$ where $k$ is a positive integer, and $N$ is a prime, and $m\ (\gt 1)$ is an integer satisfying $\gcd(m,N)=1$, then $n$ is divisible by a prime $p$ such that $p|2^{N^{k}\gcd(m,\frac{N-1}{2})}+1$.
Proof : Let $p$ be a prime such that $p|m$. Then $2^{2n}\equiv 1\pmod{pN^k},2^{(p-1)(N-1)N^{k-1}}\equiv 1 \pmod{pN^k}$ (Euler’s Theorem), so $2^{2\gcd(n,\ (p-1)\frac{N-1}{2}N^{k-1}\ )} \equiv 1 \pmod{pN^k}$,$$2^{2N^{k−1}\ \gcd(mN,\ (p-1)\frac{N-1}{2})}\equiv 1 \pmod p.$$ Now, we have $\gcd(mN, (p-1)\frac{N-1}{2})=\gcd(m,\frac{N-1}{2})$ or $N\gcd(m,\frac{N-1}{2})$. If $\gcd(mN, (p-1)\frac{N-1}{2})=\gcd(m,\frac{N-1}{2})$, then we get $2^{2N^{k−1}\gcd(m,\frac{N-1}{2})}\equiv 1 \pmod p.$ If $\gcd(mN, (p-1)\frac{N-1}{2})= N\gcd(m,\frac{N-1}{2})$, then we get $2^{2N^{k}\gcd(m,\frac{N-1}{2})}\equiv 1 \pmod p$, so the latter congruence certainly holds. So, we have $(2^{N^{k}\gcd(m,\frac{N-1}{2})}+1)(2^{N^{k}\gcd(m,\frac{N-1}{2})}-1)\equiv 0\pmod p$. Since $2^{N^{k}\gcd(m,\frac{N-1}{2})}-1\not\equiv 0 \pmod p$, we have $2^{N^{k}\gcd(m,\frac{N-1}{2})}+1\equiv 0\pmod p$.$\quad\blacksquare$