A generalization of Feit–Thompson conjecture, for square-free integers

I asked the following question with my account that I have for these sites Mathematics Stack Exchange and MathOverflow. The bounty that I offered in MSE expired without answers. The post that I refer is this MSE 3591614 (March, 23th 2020). See the useful comments that were added in MSE.

Few weeks ago I wondered about if the following conjecture is in the literature or well if it is possible to find a counterexample. I evoke a generalization of a well-known conjecture, I mean the Feit–Thompson conjecture from Wikipedia.

An integer $$n>1$$ is square-free or squarefree if has no repeated prime factors. For example $$n=15$$ is squarefree, while than $$12$$ has repeated prime factors.

Conjecture. There are no distinct square-free integers $$p$$ and $$q$$ (both greater than $$1$$) with $$\gcd(p,q)=1$$ such that $$\frac{p^q-1}{p-1}$$ divides $$\frac{q^p-1}{q-1}$$.

Question. What work can be done about the veracity of this conjecture? Since Feit–Thompson conjecture seems a particular case of this, then I'm asking if previous conjecture is in the literature, or well if you've some approach or heuristic about the Conjecture or well if it is possible to find a counterexample. Many thanks.

I wrote a Pari/GP program and I've tested the Conjecture, let's say for the segments of integers $$2\leq p,q\leq 2000$$. I hope that there were not mistakes and this post is interesting.

References:

[1] Feit, W. and Thompson, J. G., A Solvability Criterion for Finite Groups and Some Consequences, Proc. Nat. Acad. Sci. USA 48, 968-970, (1962).

• To the moderator team of MO/MSE, as soon there is feedback about if this question is interesting for MathOverflow, feel free to handle the MSE 3591614 question as you consider it is right (I say this since I never had a cross-posted question). – user142929 Apr 11 '20 at 9:35
• Maybe this could be reduced to the original Feit-Thompson conjecture considering that, writing the set of prime factors of $p$ as $p_i$ and the set of prime factors of $q$ as $q_j$ one has $(p_i,q_j)=1$. The only idea I have so far is to consider automorphisms of the rings of $p$-adic and $q$-adic integers. – Sylvain JULIEN Apr 11 '20 at 15:49
• More precisely, maybe we can try to prove that FTC is equivalent to $p\neq q\Longleftrightarrow Aut(Z_{p})\cap Aut(Z_{q})=\{e\}$. – Sylvain JULIEN Apr 11 '20 at 16:06
• just reaching out to you when I read you are losing your internet access from home; we will miss you as a frequent contributor! – Carlo Beenakker Nov 11 '20 at 9:57
• Thank you very much, I appreciate this community of users very much @CarloBeenakker , I hope your colleagues (professors) and users of this website are in good health. – user142929 Nov 12 '20 at 16:30