This is literaly Schur Theorem for $P(k)=k^3+6$

In this special case, you can also just notice that any prime $p \equiv 2\pmod{3}$ is in $P$ since for these primes, $k \mapsto k^3$ is bijective modulo $p$.

I think your problem is the particular case of this theorem : If $ T $ is polynomial, then its gets infinite prime divisers. For demonstration, we use the method of Euclid to demonstrate the infinity of prime numbers. By writting : $ T_a(x) = a_nx^n +a_{n-1}x^{n-1}+.....+a_1x+a_0 $ $a_0,a_1....a_{n-1}$ integers; $a_n$ a non zero integer and $x$ integer variable. Let suppose by contradiction that there exists finite prime numbers p_1, p_2,........p_n and let $$p_n!! = p_1.p_2.p_3......p_{n-1}.p_n$$ We distinguich 3 cases: - if $ a_0 = 1$ it is clear that no one of $p_i$ don't divide $T_1(p_n!!)$ wchich is a contradiction (Because of the fondamental of arithmetic and $|T_1(p_n!!)| > p_n$) -if $|a_0| > 1$, we get: $$T_a(a_0.p_n!!) = a_0. T_1(p_n!!)$$ therefore we get the same contradiction as below - if $a_0. = 0$, it remain to study the last two cases for the firs $a_1,a_2....a_{n-1}$ that is non zero and the contradiction is clear. In conclusion, any polynomial of integer coefficient and the first coefficient a non zero is divided by infinite prime numbers.

edemafa wrote: ... - if $a_0. = 0$, it remain to study the last two cases for the firs $a_1,a_2....a_{n-1}$ that is non zero and the contradiction is clear. Observation. For the case $ a-0= 0 $ we can say that for every integer $k, k | T_0(k) $ , since we get infinte prime number which are all integers, the proposition is true.