I have seen the past threads but I think I have another proof, though am not entirely convinced.
Suppose there are only finitely many primes $p_1, ..., p_n$ of the form $6k-1$ and then consider the number
$N = (p_1.p_2. ... .p_n)^2 - 1$
If all of the primes dividing N are of the form $6k+1$ then $N$ leaves remainder $1$ upon division by $6$. However, the RHS has remainder $1-1 = 0$ or $-1-1 = -2$ upon division by $6$, depending on whether $n$ is even or odd.
Therefore there must be at least one prime of the form $6k-1$ dividing $N$, and thus some $p_j$ divides N, and so divides $(p_1.p_2. ... .p_n)^2 - N$ which is $1$, a contradiction.
Apologies for this question if it appears unnecessary I'm just beginning number theory and trying to get to grips with it.