Is it possible for a natural number $k$ to exist such that for all primes $p$: $$(p+1)k-1$$ is composite?
(i.e., can all $3k-1$, $4k-1$, $6k-1$, $8k-1$, $12k-1$, $14k-1$, $18k-1$, $20k-1$, $24k-1$, $30k-1$, $32k-1$, $38k-1$, $42k-1$, $44k-1$, $48k-1$, ... be composite?)
Seva proved that Dickson's conjecture implies there is no such $k$. One only needs a weak form of this conjecture. Such a weak form could potentially be much easier than for example the problem on infinitely Sophie Germain primes, as one does not need to know whether for two fixed forms, such as $f_1(n)=n$ and $f_2(n)=kn+(k-1)$ are prime simulanteously, infinitely often, but just needs to know they produce one prime pair.
One can prove the following: For every integer $k\geq 2$, there exists some power $k^r$ (where $r$ may depend on $k$) such that there exists some prime $p$ with $k^r(p+1)-1$ also prime. (This statement does not rule out the possibilty that some $k$ could exist. for which all $k(+1)-1$ are composite, but it illustrates the view that some unconditional version of a weak form of Dickson's-type conjectures could potentially solve the problem.
Proof: Results of Maynard-Tao (see e.g. James Maynard: Dense clusters of primes in subsets, https://arxiv.org/abs/1405.2593) imply that two of the polynomials $f_i(t)=k^i t-1$ take prime values, simultaneously, infinitely often. (Actually, we do not even need "infinitely often", we just need one such prime pair.) With $p_1=k^i t-1, p_2=(p_1+1)k^{j-i}-1=k^j t-1$. there exists a pair of primes $p, k^r (p+1)-1$. (We do not have good control over $r=j-i$ but that does not matter. Some upper bound on $r$ is possible in view of the quantitatve nature of Maynard's results).
Hence for $k^r$ the expression $k^r(p+1)-1$ is not always composite.
In view of quantitatve versions of Linnik's theorem on primes in progression,
or Dirichlet's theorem:
for every $p+1$ there are many values $k$ making $(p+1)k-1$ a prime.
In this way one could sieve out the possible values for $k\leq N$ in an interval.
Probabilistically,
if one uses more than about $(\log N)^2$ linear forms, then
there should be very few $k \leq N$ left.
This is, as $(1- \frac{1}{\log N})^{(\log N)^2}\approx \frac{1}{N}$.
As Seve remarked that one can study $k=\frac{p+1}{q+1}$. This topic has been extensively studied by Peter Elliott. (Mathscinet: "Elliott" combined with "rationals" gives many papers).
In particular: The multiplicative group of rationals generated by the shifted primes, I. by P.D.T.A. Elliott, Journal für die reine und angewandte Mathematik (1995), 463, page 169-216. https://eudml.org/doc/153727
Here Elliott explicitly conjectures: (Conjecture 1): Every positive rational $r$ has a representation $r=\frac{p+1}{q+1}$, $p$ and $q$ primes.
He comments that the strengthening has infinitely many representations is an analogue to Schinzel's hypothesis, and would follow from Dickson's conjectures. He studies the group of rationals generated by these shifted primes and achieves many deep results.
Most certainly such $k$ do not exist. I am not sure whether there is a chance to prove this unconditionally, as even the situation with Sophie Germain primes (which are primes $p$ such that $2p+1$ is also prime, related to the case $k=2$ of your question) is unclear.
Conditionally, the non-existence of $k$ is an immediate consequence of Dickson's conjecture stating that for a finite set of linear polynomials $f_1(n)=a_1n+b_1,\dotsc,f_m(n)=a_mn + b_m$ with integer coefficients and $a_1,\dotsc,a_m\ge 1$, there are infinitely many positive integers $n$ for which the values of these polynomials are all prime, unless the product $f_1(n)\dotsb f_m(n)$ has a fixed prime factor. For your problem, take $m=2$, $f_1(n)=n$, and $f_2(n)=kn+(k-1)$.
A nice reformulation of your question is as follows: is it true that every integer $k>0$ is representable as $$ k = \frac{p+1}{q+1} $$ with $p$ and $q$ both prime? Dickson's conjecture implies that, in fact, any rational $k>0$ has infinitely many such representations: write $k=u/v$ and consider the polynomials $f_1(n)=un-1$ and $f_2(n)=vn-1$.
This is a comment on possible path to an unconditional proof.
If that were true you would have: $(p_n+1)k-1=a_nb_n$, where $1<a_n,b_n<(p_n+1)k$, that is, you would have $k=\frac {a_n b_n+1}{p_n+1}$ for every $n \in\mathbb N$ which implies $\frac{1}{k}=\frac{p_n+1}{a_n b_n+1}$ and that implies $$\sum_{l=1}^{+ \infty}(\frac{1}{k})^l=\sum_{l=1}^{+ \infty}(\frac{p_{l}+1}{a_l b_l+1})^l$$.
On the left hand side there is a rational number and I would bet that on the right hand side we have an irrational.
I'm sorry to say this, but I just proved that the above is indirectly an unsolved problem. It is Dickson's Conjecture when $k=2$. I suggest to not work too much on this. Thank you for all the insights and ideas.
