All positive Pell solutions are powers of the fundamental solution

Because ${\epsilon }_D>1$, the positive powers $\alpha ,\alpha {\epsilon }_D^{-1},\alpha {\epsilon }_D^{-2},\dots$ form a strictly decreasing sequence. By [F2], each term is again an integral norm-one element. If $\beta =u+v\sqrt{D}$ is any integral norm-one element with $\beta >1$, then $0<{\beta }^{-1}=u-v\sqrt{D}<1,$ so $u=\frac{\beta +{\beta }^{-1}}{2},v=\frac{\beta -{\beta }^{-1}}{2\sqrt{D}}>0.$ Thus every term $>1$ in the displayed sequence is again a positive Pell solution. Their $x$-coordinates are positive integers and strictly decrease, because $1<\gamma <\delta$ with $N_D(\gamma )=N_D(\delta )=1$ implies ${\gamma }^{-1}>{\delta }^{-1}>0,\gamma +{\gamma }^{-1}<\delta +{\delta }^{-1},$ hence the corresponding $x$-coordinates satisfy $x(\gamma )<x(\delta )$. Therefore only finitely many indices $j\ge 0$ satisfy $\alpha {\epsilon }_D^{-j}>1$. The set $K:=\{j\in {\mathbb{Z}}_{\ge 0}:\alpha {\epsilon }_D^{-j}\ge 1\}$ is nonempty because $0\in K$, so it has a greatest element $k$. Then $1\le \alpha {\epsilon }_D^{-k},$ and maximality of $k$ gives $\alpha {\epsilon }_D^{-(k+1)}<1.$ Multiplying the last inequality by ${\epsilon }_D$ yields $\alpha {\epsilon }_D^{-k}<{\epsilon }_D,$ so $1\le \alpha {\epsilon }_D^{-k}<{\epsilon }_D\mathrm{.}$

Put $\beta :=\alpha {\epsilon }_D^{-k}\mathrm{.}$ Step 1.1 and [F2] show that $\beta$ is an integral norm-one element with $1\le \beta <{\epsilon }_D$. If $\beta >1$, then step 1.1 shows that $\beta$ is a positive Pell solution with $x$-coordinate smaller than $x_1$, contradicting [F1]. Hence $\beta =1$, so $\alpha ={\epsilon }_D^k\mathrm{.}$ Because $\alpha >1$, the exponent cannot be $0$, so $k\ge 1$.

If also $\alpha ={\epsilon }_D^m={\epsilon }_D^n$ with $m<n$, then multiplying by ${\epsilon }_D^{-m}$ inside the group from [F2] gives $1={\epsilon }_D^{n-m},$ impossible because ${\epsilon }_D>1$. So the exponent of a positive solution is unique.