Here's another proof that may be of use. Let's go back to the real situation, and consider the function $\displaystyle g(r ):= {1+r^{n+1}\over 2} $. It maps the unit interval $[0,1]$ into itself. The numbers $\beta:=1/\alpha$ and $1$ are its fixed points. In fact, $g(r )> r$ in $[0,\beta)$ and $g(r )< r$ in $[\beta,1)$, and $g(1)=1$. Since $g$ is increasing (and continuous) this immediately implies that for any $r\in [0,1)$ the iterates $g^k(r )$ converge monotonically to $\beta$. Finally, note that $0<g'(\beta)<1$, so by convexity $g$ is a contraction in a whole interval $[0,\rho]$ for some $\rho>\beta$.
Now complexify, and note that the map $\displaystyle g(z ):= {1+z^{n+1}\over 2} $ verifies $|g(z)|\le g(|z|)$ and $|g'(z)|=g'(|z|)$. As a consequence, for any $\beta\le r<1$ the closed disk of radius $r$ is a $g$-invariant set, and it is mapped into the disk of radius $\rho$ after finitely many iterations of $g$ (how many, depending on $r$). As a consequence, any fixed point of $g$ on the open unit disk must belong to the closed disk of radius $\rho$, where $g$ is a contraction. We conclude that $z=\beta$ is the unique solution to $2z=1+z^{n+1}$ with $|z|<1$. Moreover, $g$ maps the closed unit disk in the disk of radius $1/2$ centered at $1/2$, whose only unit norm point is $z=1$, which is therefore the only fixed point of unit norm.
Passing to the reciprocal equation: the only solutions to $2x^n=x^{n+1}+1$ with $|x|\ge1$ are $x=\alpha$ and $x=1$, whence your thesis.