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):
- 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".