For a prime $p$, let $M_p = 2^p-1$ be a (Mersenne) number with exactly two prime divisors, and let $P(p)$ be the largest of these two. Clearly $P(p) > \sqrt{M_p}$. This is very likely a hard question, but can we say anything about how far $P(p)$ is from $\sqrt{M_p}$? For instance, could it possibly be true that the ratio $$ R(p) = \dfrac{P(p)}{\sqrt{M_p}} $$ is unbounded?
Murata and Pomerance conjectured in
https://math.dartmouth.edu/~carlp/PDF/murata4.pdf
that $P(p)$ grows asymptotically faster than any polynomial in $p$. But this is still under $\sqrt{M_p}$, not useful in our case of $P(p) >\sqrt{M_p} $ with $M_p$ having two prime factors.
Given that all the prime factors of $M_p$ are congruent to $1$ modulo $2p$, one initial idea could be to get an asymptotic expression for the smallest prime greater than $\sqrt{M_p}$ and which is congruent to $1$ modulo $2p$. Can this be feasible? Or perhaps there is another way to go about this, if any at all? Thank you very much.
Here's some data. The pairs are $(p, R(p))$ for which $M_p$ has exactly two prime divisors. Notice the large values of $R(p)$ towards the end. Although there are some proportionally smaller ones, as the ones for $p = 11, 101, 137, 149$. But even here $R(p) > p^{1/4}$. Curiously, these last primes $p$ are also the first entries of twin prime pairs (what??).
(11, 1.9671210479762813)
(23, 61.623600066097914)
(37, 1662.455605799928)
(41, 110.9381611714366)
(59, 4219.204811276472)
(67, 62.713050038434965)
(83, 18622086898.05674)
(97, 34774677167.19322)
(101, 214.23442518117758)
(103, 1248743.6542854793)
(109, 34150923.513927594)
(131, 1.9838506198224765e+17)
(137, 13.03069933308962)
(139, 148389420.30816036)
(149, 308.27243485390295)
(167, 5.82260553430773e+18)
(197, 5.986138210208216e+25)
(199, 5.448860357248897e+18)
(227, 5.4420261679031904e+17)
(241, 8.5444423288905e+28)
(269, 2.2282001358749983e+33)
(271, 4.041186699015196e+30)