If $p_1$ and $p_2$ are primes, prove that there exists an integer such that $p_1+k$ is a prime but $p_2+k$ is not.

Note that the numbers $P_1+nP_2, n\in \mathbb N$ are in arithmetic progression and this contains (by Dirichlet) infinitely many primes. So choose $n\ge 1$ which makes this a prime. $P_2+nP_2$ is clearly not prime.

Here is a more elementary approach than just quoting Dirichlet.

Suppose $P_2\gt P_1$ and $P_2-P_1=d$. If $P_3=P_1+k$ we want $P_3+d=P_2+k$ not to be a prime.

So given $d$ we need to find a prime $P_3$ with $P_3+d$ not a prime.

Now we know that there are arbitrarily large gaps between primes e.g. $n!+2, \dots n!+n$ gives a gap of size at least $n-1$. (all these are composite by construction, so the prime before and the prime after will have a big gap)

We can choose $n\gt d+1$ here, so there will be a $P_3$ distance greater than $d$ from the next prime, and we can choose $k$ to pick this out.

Assume wlog $p_1 < p_2$ and $p_1+k$ is prime iff $p_2+k$ is prime. Set $z:=p_2-p_1$. By induction, $p_1+nz$ is prime for all $n$. This is impossible since the primes have density zero.