In general, you can find the minimum polynomial for an algebraic number $\alpha$ by determining the smallest power of $\alpha$ for which $\{1,\alpha,\alpha^2,\ldots,\alpha^n\}$ is linearly dependent over $\mathbb{Q}$.
For example, let $\alpha=\sqrt{2} + \sqrt{3}$. Then
\begin{align*}
\alpha^2 &= 5+2\sqrt{6} \\\
\alpha^3 &= 11\sqrt{2}+9\sqrt{3} \\\
\alpha^4 &= 49+20\sqrt{6}
\end{align*}
Since $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ are linearly independent over $\mathbb{Q}$, we can think of the powers of $\alpha$ as vectors:
$$
1 = \begin{bmatrix}1\\\ 0\\\ 0\\\ 0\end{bmatrix},
\qquad
\alpha = \begin{bmatrix}0\\\ 1\\\ 1\\\ 0\end{bmatrix},
\qquad
\alpha^2 = \begin{bmatrix}5\\\ 0\\\ 0\\\ 2\end{bmatrix},
\qquad
\alpha^3 = \begin{bmatrix}0\\\ 11\\\ 9\\\ 0\end{bmatrix},
\qquad
\alpha^4 = \begin{bmatrix}49\\\ 0\\\ 0\\\ 20\end{bmatrix}
$$
As you can see, $\{1,\alpha,\alpha^2,\alpha^3\}$ is linearly independent, so $\alpha$ is not the root of any cubic polynomial. However, $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ is linearly dependent, with
$$
\alpha^4 - 10\alpha^2 + 1 \;=\; 0
$$
It follows that $x^4-10x^2+1$ is the minimum polynomial for $\alpha$.
This technique depends on being able to recognize a useful set of linearly independent elements like $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$. Depending on how much Galois theory you know, it may be hard to prove that elements like this are linearly independent over $\mathbb{Q}$. In this case, you can use this technique to guess the minimum polynomial, and then prove that you are correct by proving that the polynomial you found is irreducible.
For example, if $\alpha = i\sqrt[4]{2}$, then
$$
\alpha^2 = -\sqrt{2},\qquad \alpha^3 = -i\sqrt[4]{8},\qquad \alpha^4 = 2.
$$
It seems clear that $ \{1,\alpha,\alpha^2,\alpha^3\} $ is linearly independent, while $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ satisfies $\alpha^4-2 = 0$. Thus $x^4-2$ should indeed be the minimum polynomial over $ \mathbb{Q} $, though the easiest way to prove it is to show that $x^4 -2$ is irreducible. This can be done using Eisenstein's criterion, or by checking that $ x^4-2 $ has no roots and does not factor into quadratics modulo $4$.
Best Answer
Let's look for factors of degree $3$. Suppose (over $\mathbb F_2$) $$x^6 + x^3 + 1 = (x^3 + a_2 x^2 + a_1 x + a_0)(x^3 + b_2 x^2 + b_1 x + b_0)$$ Since $a_0 b_0 = 1$, we must have $a_0 = b_0 = 1$. By looking at the $x^5$ and $x^1$ terms, you must have $a_2=b_2$ and $a_1=b_1$. But then the $x^3$ term on the right is $0$.
The case of a factor of degree $2$ is similar.