Solution: Note that \begin{align*} x^4 + x^2 + 1 & = x^4 + 2x^2 +1 - x^2 \\ & = \left( x^2 + 1 \right) ^2 - x^2 \\ & = \left( x^2 - x + 1 \right) \left( x^2 + x + 1 \right) . \end{align*} Therefore, $x^4 + x^2 + 1$ is reducible over $\mathbb{Q} $. The zeros of the given polynomials are the zeros of its factors. The zeros of $x^2 \pm x + 1$ are \[ \frac{\pm 1 \pm \sqrt{1 - 4} }{2} = \frac{\pm 1 \pm \iota \sqrt{3} }{2}, \] where $\iota ^2 = -1$. Therefore, $\mathbb{Q} \left( \iota \sqrt{3} \right) $ contains all zeros of the polynomial $x^4 + x^2 + 1$. Therefore, the splitting field $\mathbb{F} $ of $x^4 + x^2 + 1$ is a subfield of $\mathbb{Q} \left( \iota \sqrt{3} \right) $. Since $\mathbb{Q} $ does not contain the roots, so $\mathbb{F} \neq \mathbb{Q} $.

Since the minimal polynomial of $\iota \sqrt{3}$ is $x^2 + 3$, so the degree of $\mathbb{Q} \left( \iota \sqrt{3} \right) $ over $\mathbb{Q} $ is $2$. Therefore, $\mathbb{F} = \mathbb{Q} \left( \iota \sqrt{3} \right) $.