Nothing wrong Robert Israel's answer. I simply cannot resist describing a bit of theory leading to some (but not all) solutions.
TL;DR; version: If $\beta$ is a primitive element of $GF(256)$, then
$$
x^8+\beta^{85}x^2+x
$$
has eight distinct zeros in $GF(256)$, and for them we have $e_1=e_2=e_3=e_4=e_5=0.$ On with the longer version...
Assume that $V\subset GF(2^q)$ is a 3-dimensional subspace (over $GF(2)$). Consider the octic polynomial that has zeros at the points of $V$:
$$
L_V(x)=\prod_{z\in V}(x-z).
$$
I will show that $L_v$ has only terms of degrees $1,2,4$ and $8$.
Let $v\in GF(2^q)$. The polynomial
$$p_v(x)=x^2-vx$$
obviously vanishes at $0$ and $v$. Furthermore, $p_v(x)$ is a so called
linearized polynomial, meaning that
$$
p_v(x+y)=p_v(x)+p_v(y).
$$
Let $v_1,v_2,v_3$ be a basis of $V$. Because $p_{v_1}(x)$ is linearized, we see that
$$p_{v_1}(v_2)=p_{v_1}(v_2)+p_{v_1}(v_1)=p_{v_1}(v_1+v_2).$$
Therefore the composition
$$
p_{v_1,v_2}(x):=p_{p_{v_1}(v_2)}(p_{v_1}(x))
$$
vanishes at the points $0,v_1,v_2,v_1+v_2$. Furthermore, by the composition rule (and Freshman's dream in characteristic two) we see that
$p_{v_1,v_2}(x)$ only has terms of degrees $1,2$ and $4$, and hence it, too,
is a linearized polynomial.
Repeating the dose then shows that
$$
p_{v_1,v_2,v_3}(x):=p_{p_{v_1,v_2}(v_3)}(p_{v_1,v_2}(x))
$$
vanishes at the points of $V$, only has terms of degrees $1,2,4$ and $8$, and hence is also linearized. Because this octic is also monic, it must be equal to $L_V(x)$.
Anyway,
$$
L_V(x)=x^8+e_4(V)x^4+e_6(V)x^2+e_7(V)x,
$$
where the $e_i$:s are the elementary symmetric polynomials of the set $V$.
In particular, we automatically have $e_1(V)=e_2(V)=e_3(V)=0$ hopefully explaining why I wanted to explain this to you.
So selecting your elements $\alpha_i, i=1,2,\ldots,8,$ to be the elements of a subspace $V$ gives us $e_1=e_2=e_3=0$ automatically leaving us to deal with $e_4$. We are lead to look for subspaces $V$ such that
$$
L_V(x)=x^8+ Ax^2+Bx,
$$
when we will in addition to the above get $e_4=e_5=0$. To make the search simpler let us consider the effect of scaling. If we multiply the subspace $V$ by an element $\alpha\in GF(2^q)$, we get the subspace
$$
\alpha V=\{\alpha z\mid z\in V\}.
$$
Clearly $e_6(\alpha V)=\alpha^6 e_6(V)$ and $e_7(\alpha V)=\alpha^7e_7(V)$. As raising to seventh power is a bijection of $GF(2^8)$ we can, by selecting a suitable $\alpha$, without loss of generality assume that $e_7(V)=1$.
We are thus left with polynomials of the form
$$
L_a(x)=x^8+ax^2+x.
$$
We need to select $a$ in such a way that $L_a(x)$ has eight zeros in $GF(2^8)$. Because all the elements of $GF(256)$ are zeros of $P(x)=x^{256}-x$, we want to check that $L_a(x)$ is a factor of $P(x)$.
Using square-and-multiply (and Freshman's dream!!!) it is actually a pencil-and-paper calculation to find the remainder! It turns out that
$$
P(x)\equiv(1+a^{42})x^4+(a^5+a^{33}+a^{40})x^2+(1+a^4+a^{32})x\pmod{L_a(x)}.
$$
So $L_a(x)$ will give us what we want if all those three polynomials on $a$ vanish.
It follows from Bill Dubuque's answer to another question that if $a$ is a primitive third root of unity, i.e. it satisfies the equation $a^2+a+1$, then we, indeed, get
$$
L_a(x)\mid P(x).
$$
Observe that already the field $GF(4)\subset GF(256)$ contains the primitive third roots of unity. If $\beta$ is a primitive element of the field $GF(256)$, then $\beta^{85}$ can serve in the role of $a$ in $L_a(x)$.
Furthermore, by linearity of $L_a(x)$, the cosets of the subspace
$V:=\operatorname{Ker}(L_a)$ are zeros of polynomials of the form $L_a(x)+c$
for some $c\in GF(256)$. There are $32$ such cosets, so you get $8$ zeros in $GF(256)$ for $32$ different choices of $c$. Any of those $32$ cosets will
serve as your set $\alpha_i, i=1,2,\ldots,8$.
