I am trying to show that x^2-x+1 is irreducible in $\mathbb{Z_p}$ iff p $\equiv$ 2 mod 3
I'm not sure if i'm approaching this right but I think that if $x^2-x+1$ had a linear factor, $x^2-x+1=0$ would have a root (say a) in $\mathbb{Z_p}$
Using the fact that $(x^3-1)=(x+1)(x^2-x+1)$ which implies $a^3=-1$ so $a \neq 1$ but $a^6=1$ and a has order 3 (not entirely sure )
I know the order of $\mathbb{Z_p} = p-1$ and I think $p\equiv$ 2 mod 3 means there is no such element in $\mathbb{Z_p}$ with order 3, so the polynomial is irreducible
This could be completely wrong so any help is very much appreciated, especially if there is an easier way
You have the right idea, but you got some signs wrong: $(x^3 + 1) = (x+1)(x^2 - x + 1)$.
So $x^2 - x + 1$ has a linear factor iff there exists an $a \in \mathbb Z_p$ such that $a^3 = -1$ but $a \neq -1$.
Another fact that you may find useful is that the multiplicative group $\mathbb Z_p^\times$ is a cyclic group. So there exists a generator $g \in \mathbb Z_p^{\times}$, and you can write any $a \in \mathbb Z_p^\times$ in the form $a = g^e$ for some integer $e$.
The order of $\mathbb Z_p^\times$ is $p-1$, and $(-1)^2 = 1$. So ask yourself: for what values of $e$ does $g^e = -1$? This should put you on the right track.