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, $(1- \frac{1}{\log N})^{(\log N)^2}\approx \frac{1}{N}$.

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.

Seva proved that Dickson's conjecture implies there is no such $k$, and commented: unconditionally "there is no chance". It is actually less hopeless as one only needs a very weak form of the conjecture.

Let $k\geq 2$, (the question makes no sense for $k=1$).

Let $Q(k)$ be the statement: "For all primes $p$ the value $(p+1)k−1 is composite."

Step 1) If $Q(k)$ is true, then it is also true for all multiples of $k$, in particular, then $Q(k^r)$ is true, for all positive integers $r$.

This is obvious as one only looks at a subset of the conditions required for $Q(k)$.

Step 2) For any integer $k\geq 2$, there exist some power $k^r$ (where $r$ may depend on $k$) such that $Q(k^r)$ is not true.

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, and so $Q(k^r)$ is not true.

Step 3):

1. and 2) are in contradiction, hence the assumption of 1) that $Q(k)$ is true can never be satisfied.

The answer to your question is "No".

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, $(1- \frac{1}{\log N})^{(\log N)^2}\approx \frac{1}{N}$.