Solutions of $z^6 + 1 = 0$ Solve:
$$z^6 + 1 = 0$$
That lie in the top region of the plane.
We know that:
$$(z^2 + 1)(z^4 - z^2 + 1) = 0$$
$$z = -i, i$$ 
We need to solve:
$$((z^2)^2 - (z)^2 + 1) = 0$$
$$z = \frac{1 \pm \sqrt{-3}}{2}$$
But this is incorrect. How to do this then?

The simplest way of solving this equation is the method based on DeMoivre's Formula that Lab Bhattacharjee outlined.  
That said, you can make your method work.  You found the roots $z \pm i$ by setting the factor $z^2 + 1$ equal to zero.  As Rasolnikov and 5xum noted, you should have obtained 
$$z^2 = \frac{1 \pm \sqrt{-3}}{2}$$
when you set the factor $z^4 - z^2 + 1$ equal to zero.
Let $z = a + bi$, with $a, b \in \mathbb{R}$.  Then 
\begin{align*}
z^2 & = \frac{1 \pm i\sqrt{3}}{2}\\
(a + bi)^2 & = \frac{1 \pm i\sqrt{3}}{2}\\
a^2 + 2abi - b^2 & = \frac{1 \pm i\sqrt{3}}{2}
\end{align*}
Equating real and imaginary parts yields
\begin{align*}
a^2 - b^2 & = \frac{1}{2}\tag{1}\\
2ab & = \pm\frac{\sqrt{3}}{2}\tag{2}
\end{align*}
Solving equation 2 for $b$ yields
$$b = \frac{\pm\sqrt{3}}{4a}\tag{3}$$
Substituting this expression in equation 1 yields
\begin{align*}
a^2 - \frac{3}{16a^2} & = \frac{1}{2}\\
16a^4 - 3 & = 8a^2\\
16a^4 - 8a^2 & = 3\\
16a^4 - 8a^2 + 1 & = 4 && \text{complete the square}\\
(4a^2 - 1)^2 & = 4\\
4a^2 - 1 & = \pm 2\\
4a^2 & = 3 && \text{since $a \in \mathbb{R}$}\\
a^2 & = \frac{3}{4}\\
a & = \pm \frac{\sqrt{3}}{2}
\end{align*}
Substituting this expression into equation 3 yields the four roots 
$$z = \pm\frac{\sqrt{3}}{2} \pm \frac{1}{2}i$$
of the equation $z^4 - z^2 + 1 = 0$. As you can check, these roots correspond to the values $n = 0, 2, 3, 5$ in the formula Lab provided.   
