You've got something wrong: the roots of $t^2+t+1$ are the complex cubic roots of one, not of $-1$: $t^3-1 = (t-1)(t^2+t+1)$, so every root of $t^2+t+1$ satisfies $\alpha^3=1$). That means that you actually want the cubic roots of some of the cubic roots of $1$; that is, you want some ninth roots of $1$ (not of $-1$).
Note that
$$(x^6+x^3+1)(x-1)(x^2+x+1) = x^9-1.$$
So the roots of $x^6+x^3+1$ are all ninth roots of $1$. Moreover, those ninth roots should not be equal to $1$, nor be cubic roots of $1$ (the roots of $x^2+x+1$ are the nonreal cubic roots of $1$): since $x^9-1$ is relatively prime to $(x^9-1)' = 9x^8$, the polynomial $x^9-1$ has no repeated roots. So any root of $x^9-1$ is either a root of $x^6+x^3+1$, or a root of $x^2+x+1$, or a root of $x-1$, but it cannot be a root of two of them.
If $\zeta$ is a primitive ninth root of $1$ (e.g., $\zeta = e^{i2\pi/9}$), then $\zeta^k$ is also a ninth root of $1$ for all $k$; it is a cubic root of $1$ if and only if $3|k$, and it is equal to $1$ if and only if $9|k$. So the roots of $x^6+x^3+1$ are precisely $\zeta$, $\zeta^2$, $\zeta^4$, $\zeta^5$, $\zeta^7$, and $\zeta^8$. They are all contained in $\mathbb{Q}(\zeta)$, which is necessarily contained in the splitting field. Thus, the splitting field is $\mathbb{Q}(\zeta)$, where $\zeta$ is any primitive ninth root of $1$.
The splitting field is either degree $3$ or degree $6$ over $\mathbb{Z}_2$, hence it is either $\mathbb{F}_8$ or $\mathbb{F}_{64}$.
Let $\alpha$ be a root, so that $\mathbb{F}_8 = \mathbb{F}(\alpha)$. The elements are of the form $a+b\alpha+c\alpha^2$, with $\alpha^3=\alpha+1$.
Now, the question is whether any of these elements besides $\alpha$ is a root of the original polynomial $x^3+x+1$.
Note that $(\alpha^2)^3 = (\alpha^3)^2 = (\alpha+1)^2 = \alpha^2+1$, and so if we plug in $\alpha^2$ into the polynomial we have
$$\alpha^6 + \alpha^2 + 1 = \alpha^2+1+\alpha^2+1= 0.$$
Thus, $\alpha^2$ is also a root. So the polynomial has at least two roots in $\mathbb{F}_8$, and so splits there.
Best Answer
Let $\omega = e^{\pi i/3}$. The roots of $y^2 + y + 1$ are the primitive third roots of unity $\omega^2$ and $\omega^4$ (note that $y^3 - 1 = (y-1)(y^2 + y + 1)$), so the roots of $x^4 + x^2 + 1$ are the sixth roots of unity $\omega, \omega^2, \omega^4, \omega^5$. So the splitting field is $K = \mathbb{Q}(\omega)$.
The polynomial $x^4 + x^2 + 1$ factors as $(x^2 + x + 1) (x^2 - x + 1)$, so $[K: \mathbb{Q}] = 2$. (You can also see this by showing directly that $\mathbb{Q}(\omega) = \mathbb{Q}(\omega^2)$, and $\omega^2$ has $y^2 + y + 1$ as a minimal polynomial.)