How many zeros of $z^{17}+z-1$ lie in the region $|z|<1$? General problem:
Conjecture (B. Galo and Jack D'Aurizio) Let $p$ be a prime number then the number of zeros of $z^{p}+z-1$ that lie in the region $|z|<1$ is:
1)  $⌊\frac{p}{3}⌋-2$ (where $⌊*⌋$ is the lower integer function). If $ p≡5 (mod6)$ but $p\neq 5$.
2) $⌊\frac{p}{3}⌋$. If this result is odd and $p\not\equiv 5 (mod6)$
3) $⌊\frac{p}{3}⌋+1$. If $⌊\frac{p}{3}⌋$ is even number. 
I have to thanks Jack D'Aurizo for the "review" of my attempt and for sketching the proof.
 A: If $p\equiv 5\pmod{6}$ the polynomial $z^p+z-1$ has exactly two roots on the unit circle, at $\exp\left(\pm\frac{\pi i}{3}\right)$. Conversely, if $z^p+z-1$ has a root on the unit circle such root has to lie at the intersection of the sets $|z|=1$ and $|1-z|=1$, so $p\not\equiv 5\pmod{6}$ grants that $z^p+z-1$ has no root on the unit circle. By a numerical approximation of the integral
$$ \frac{1}{2\pi i}\oint_{|z|=1-\varepsilon}\frac{f'(z)}{f(z)}\,dz $$
for $f(z)=z^p+z-1$ it is not difficult to arrive at the following conjecture:

Conj. About $\frac{p}{3}$ roots of the polynomial $z^p+z-1$ lie in the region $|z|<1$.

Here it comes a convincing geometric argument for proving such a conjecture. The number of roots of $f(z)$ in the region $|z|\leq 1$ is given by the topological degree/winding number around the origin of the curve
$$\gamma(\theta) = \left( \cos(p\theta)+\cos(\theta)-1,\sin(p\theta)+\sin(\theta)\right) $$
that is a generalized Lissajous curve/ epitrochoid. For $p=17$ and $\theta\in[0,2\pi]$, for instance, it has the following structure:
$\hspace{2cm}$
hence it is essentially given by a circle with unit radius that is rotating around the point $z=-1$ in seventeen steps. By trivial trigonometry about one third of such circles actually enclose the origin, and that leads to the $\frac{p}{3}$ mentioned before. By studying the behaviour of $\gamma(\theta)$ as $\theta$ travels on intervals with length $\frac{2\pi}{p}$, and by noticing that the number of roots of $z^p+z-1$ in $|z|<1$ is clearly odd, we have that:

Theo. If $p\geq 11$ is a prime of the form $6k+1$ or $6k+5$, there are $2k+1$ roots of $q(z)=z^p+z-1$ in the region $|z|<1$. There are two roots of $q(z)$ on the unit circle, at $z=\exp\left(\pm\frac{\pi i}{3}\right)$, iff $p=6k+5$.

A: One approach would be to apply the bilinear transformation $z = (1+x)/(1-x)$, then use Routh-Hurwitz. The calculation should not be too difficult since most of the coefficients are missing.
A: Visualiztion
Start with
$$
f(z) = z^{17}+z-1
$$
The Re and Im components are shown next.

The $0$ contours, the line where
$$
\color{blue}{\text{Re } f= 0}, \qquad \color{red}{\text{Im } f= 0}
$$
are shown below against the unit circle. The roots where $f(z)=0$ are where the blue and red contours intersect. We see six roots within the unit disk.

Observation
While using the argument principle, a pattern emerges.
Given
$$
 g(k,z) = z^{k} - z + 1
$$
compute the series expansion for
$$
 \frac{g'(z)}{g(z)}
$$
The expansion coefficients are
$$
\begin{array}{lrrrrrrrrrrrrrrrrrrrr}
 k & const & w^1 & w^2 & w^3 & w^4 & w^5 & w^6 & w^7 & w^8 & w^9  & w^{10}\\\hline
 2 & -1 &  1 &  2 & \dots \\
 3 & -1 & -1 &  2 &  3 &  4 & \dots \\
 4 & -1 & -1 & -1 &  3 &  4 &  5 & 6 & \dots \\
 5 & -1 & -1 & -1 & -1 &  4 &  5 & 6 & 7 & 8 & \dots \\
 6 & -1 & -1 & -1 & -1 & -1 &  5 & 6 & 7 & 8 & \dots \\
 7 & -1 & -1 & -1 & -1 & -1 & -1 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \dots \\
\end{array}
$$
