Primitive elements of GF(8)

I'm trying to find the primitive elements of $GF(8),$ the minimal polynomials of all elements of $GF(8)$ and their roots, and calculate the powers of $\alpha^i$ for $x^3 + x + 1.$

If I did my math correct, I found the minimal polynomials to be $x, x + 1, x^3 + x + 1,$ and $x^3 + x^2 + 1,$ and the primitive elements to be $\alpha, \dots, \alpha^6 $

Would the powers of $\alpha^i$ as a polynomial (of degree at most two) be: $\alpha, \alpha^2, \alpha+ 1, \alpha^2 + \alpha, \alpha^2 + \alpha + 1,$ and $\alpha^2 + 1$?

Am I on the right track?

Those are all correct. Here's everything presented in a table:

$$\begin{array}{lll} \textbf{element} & \textbf{reduced} & \textbf{min poly} \\ 0 & 0 & x \\ \alpha^0 & 1 & x+1 \\ \alpha^1 & \alpha & x^3+x+1 \\ \alpha^2 & \alpha^2 & x^3+x+1 \\ \alpha^3 & \alpha+1 & x^3+x^2+1 \\ \alpha^4 & \alpha^2+\alpha & x^3+x+1 \\ \alpha^5 & \alpha^2+\alpha+1 & x^3 + x^2 + 1 \\ \alpha^6 & \alpha^2+1 & x^3 + x^2 + 1 \\ \end{array}$$

What you have is correct. Some general facts that you can use to check that you have all of the polynomials are the following:

Let $p$ be prime and $q=p^k$.

$\textbf{Fact 1:}$ In $GF(q)$ the product of all monic irreducible polynomials of degree dividing $n$ is $x^{q^n}-x$.

$\textbf{Fact 2:}$ The minimal polynomial of any element of $GF(q)$ is a monic irreducible polynomial over $GF(p)$ whose degree divides $k$.

Set $p=2$, $k=3$. Then $q = 8$.

Using fact 2, we see that the minimal polynomial of any element of $GF(8)$ is a monic irreducible polynomial over $GF(2)$ whose degree is $1$ or $3$.

Using fact 1, we see that the product of all monic irreducible polynomials of degree $1$ or $3$ over $GF(2)$ is $x^8-x$.

You can check that the product of the set of polynomials you listed is $x^8-x$ over $GF(2)$.