Most certainly such $k$ does *not* exist, but there is no chance to prove this unconditionally, as we cannot even prove that there are infinitely many [Sophie Germain primes][2] (which are primes $p$ such that $2p+1$ is also prime, corresponding to the case $k=2$ of your question). 

Conditionally, the non-existence of $k$ is an immediate consequence of the [Dickson's conjecture][1] 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 forms 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 $f_2(n)=kn+(k-1)$.

[1]:https://en.wikipedia.org/wiki/Dickson%27s_conjecture
[2]:https://en.wikipedia.org/wiki/Sophie_Germain_prime