Answer: 276

Small Hint:

Divide by $z^3$ and set $w=z+z^{-1}$

Big Hint:

Factor the resulting cubic in $w$ and identify each value as $2\cos\phi$

Solution:

No root is zero. Dividing by $z^3$ and setting $w=z+z^{-1}$ gives $$\begin{aligned}w^3-2w+1&=0,\\{}(w-1)(w^2+w-1)&=0.\end{aligned}$$ Its three roots are $$\begin{aligned}1&=2\cos60^\circ,\\\frac{\sqrt5-1}{2}&=2\cos72^\circ,\\-\frac{1+\sqrt5}{2}&=2\cos144^\circ.\end{aligned}$$ For each value $w=2\cos\phi,$ the corresponding roots of the original equation are $e^{i\phi}$ and $e^{-i\phi}.$ Thus the roots with positive imaginary part have arguments $60^\circ,$ $72^\circ,$ and $144^\circ.$ Their product has argument $60+72+144=276^\circ,$ so $\theta=276.$