$$z^2+z+1=0$$ Clearly $z \not= 0$ and therefore $z+\dfrac{1}{z}=-1.$
Note that $$z^{n+1}+\frac{1}{z^{n+1}}=\left(z+\frac{1}{z}\right)\left(z^n+\frac{1}{z^n}\right)-\left(z^{n-1}+\frac{1}{z^{n-1}}\right).$$
Therefore if we define the function $U:\Bbb{N}\cup\{0\} \to \Bbb{C}$ as $$U_n=z^n+\frac{1}{z^n}$$ then we have $U(0)=2,$ $U(1)=-1$ and $$U(n+1)+U(n)+U(n-1)=0\,\,\,\,\,\,\,\,\,\forall n \in \Bbb{N}.$$
Using this recurrence you can calculate $z^n+\dfrac{1}{z^n}$ for any $n \in \Bbb{N}.$
Alos you can follow this approach.