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Here's another proof that may be of use in other similar cases.

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 $r$). Therefore 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 its unique fixed point, that 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 complex solutions to $2x^n=x^{n+1}+1$ with $|x|\ge1$ are $x=\alpha$ and $x=1$, whence your thesis.

$$*$$

Edit. Summarizing, here is a sort of minimal proof of the equivalent claim:

The only complex solutions to $2z=z^{n+1} +1$ with $|z|\le 1$ are $z=\beta:={1\over\alpha}$ and $z=1$.

(it only uses the elementary facts about the real function $g:[0,1]\ni t\mapsto{t^{n+1}+1\over2}\in [0,1]$)

Proof. Assume $\zeta$ solves $2\zeta=\zeta^{n+1} +1$ 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 identifies $r=1$. Hence $|\zeta^{n+1} +1|=2$. So $\zeta=1$.

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.

Here's another proof that may be of use in other similar cases.

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 $r$). Therefore 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 its unique fixed point, that 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 complex solutions to $2x^n=x^{n+1}+1$ with $|x|\ge1$ are $x=\alpha$ and $x=1$, whence your thesis.

$$*$$

Edit. Summarizing, here is a sort of minimal proof of the equivalent claim:

The only complex solutions to $2z=z^{n+1} +1$ with $|z|\le 1$ are $z=\beta:={1\over\alpha}$ and $z=1$.

(it only uses the elementary facts about the real function $g:[0,1]\ni t\mapsto{t^{n+1}+1\over2}\in [0,1]$)

Proof. Assume $\zeta$ solves $2\zeta=\zeta^{n+1} +1$ 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 $0<(n+1)\beta^n=g'(\beta)<1$. 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 identifies $r=1$. Hence $|\zeta^{n+1} +1|=2$. So $\zeta=1$.

Here's another proof that may be of use in other similar cases.

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 $r$). Therefore 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 its unique fixed point, that 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 complex solutions to $2x^n=x^{n+1}+1$ with $|x|\ge1$ are $x=\alpha$ and $x=1$, whence your thesis.

$$*$$

Edit. Summarizing, here is a sort of minimal proof of the equivalent claim:

The only complex solutions to $2z=z^{n+1} +1$ with $|z|\le 1$ are $z=\beta:={1\over\alpha}$ and $z=1$.

(it only uses the elementary facts about the real function $g:[0,1]\ni t\mapsto{t^{n+1}+1\over2}\in [0,1]$)

Proof. Assume $\zeta$ solves $2\zeta=\zeta^{n+1} +1$ 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 identifies $r=1$. Hence $|\zeta^{n+1} +1|=2$. So $\zeta=1$.

Here's another proof that may be of use in other similar cases.

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 $r$). Therefore 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 its unique fixed point, that 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 complex solutions to $2x^n=x^{n+1}+1$ with $|x|\ge1$ are $x=\alpha$ and $x=1$, whence your thesis.

Here's another proof that may be of use in other similar cases.

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 $r$). Therefore 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 its unique fixed point, that 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 complex solutions to $2x^n=x^{n+1}+1$ with $|x|\ge1$ are $x=\alpha$ and $x=1$, whence your thesis.

$$*$$

Edit. Summarizing, here is a sort of minimal proof of the equivalent claim:

The only complex solutions to $2z=z^{n+1} +1$ with $|z|\le 1$ are $z=\beta:={1\over\alpha}$ and $z=1$.

(it only uses the elementary facts about the real function $g:[0,1]\ni t\mapsto{t^{n+1}+1\over2}\in [0,1]$)

Proof. Assume $\zeta$ solves $2\zeta=\zeta^{n+1} +1$ 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 $0<(n+1)\beta^n=g'(\beta)<1$. 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 identifies $r=1$. Hence $|\zeta^{n+1} +1|=2$. So $\zeta=1$.