Let's prove the equivalent claim:
The only complex solutions to $2z=z^{n+1} +1$ with $|z|\le 1$ are $z={1\over\alpha}$ and $z=1$.
Remark: The number $\beta:={1\over\alpha}$ is the minimum fixed point of the increasing convex function $\displaystyle g( r):={1+r^{n+1}\over 2} $ on $\mathbb{R}_+$, and as such $0<g'(\beta)=(n+1)\beta^n<1$.
$\phantom{y}$
Proof of claim. Assume $\zeta$ verifies $2\zeta=\zeta^{n+1} +1$ and $|\zeta|=r\le 1.$
Case 1: $r \le \beta$. Then subtracting $2\beta=\beta^{n+1} +1$ we get $$2|\zeta-\beta |= |\zeta^{n+1}-\beta^{n+1}|\le |\zeta-\beta|\sum_{k=0}^nr^k\beta^{n-k}\le|\zeta-\beta |(n+1)\beta^n.$$ But $0<(n+1)\beta^n=2g'(\beta)<2$. So $\zeta=\beta$.
Case 2: $\beta< r \le1$. Then $$2r=|\zeta^{n+1} +1| \le r^{n+1} +1 \le 2r,$$ so $r^{n+1} +1 = 2r,$ and in the interval $(\beta,1]$ this forces $r=1$. Hence $|\zeta^{n+1} +1|=2$. So $\zeta^{n+1}=1$ because $|\zeta|\le1$. Then $2\zeta =\zeta^{n+1} +1=2$ and $\zeta=1.$ $\quad\square$
Passing to the reciprocal equation: the only complex solutions to $2x^n=x^{n+1}+1$ with $|x|\ge1$ are $x=\alpha$ and $x=1$, whence your thesis.