Proving $n^2 + n + 1$ has prime factors of the form 6m + 1 or 3.

Question

So here's what I did,

$n^2 + n + 1 = n(n+1)+1 = N(say)$

as either n or n+1 is even...

Hence, $N = 2r+1$

Now, take $r = 3k$ or $3k+1$ or $3k+2$

for $3k+1$, $3|N$

for $3k+2$, $N = 6k+5$, which we can break into factors of $6l+1$ and $6j+1$ till we a prime of form $6m+1$

for 3k, N = 6k+1, now I need to show that this $6k+1$ is a prime, ie $6m+1$ is a prime if $6m = (3k)(3k+1)$ OR $6m = (3k)(3k-1)$