 Solution

Use the periodic continued fraction $\sqrt{d}=[a_0;\overline{a_1,\dots ,a_m}]$. A direct matrix argument avoids any assumption about the parity of the period. LetPeriodicity says that $M$ fixes ${\theta }_1$ as a Möbius transformation, while $B$ takes ${\theta }_1$ to $\sqrt{d}$. Thus $N=\left(\begin{array}{cc}A& B_1\\ C& D\end{array}\right)$ has integer entries and fixes $\sqrt{d}$. Equating rational and irrational parts in $A\sqrt{d}+B_1=\sqrt{d}(C\sqrt{d}+D)$ gives $A=D,B_1=dC$. ConsequentlyThe eigenvalue of $M$ on $({\theta }_1,1{)}^T$ is its denominator $M_{21}{\theta }_1+M_{22}>1$; this uses $a_i\ge 1$ and ${\theta }_1>1$, including the one-term period. The corresponding eigenvalue of $N$ is $A+C\sqrt{d}>1$. If the algebraic norm is $-1$, square this unit; otherwise keep it. In either case there is a unit $u=X+Y\sqrt{d}>1$ of algebraic norm $1$ with integer $X,Y$. Its powers have integer coefficients and algebraic norm $1$, and are distinct since $u^j$ strictly increases. Hence Pell's equation has infinitely many integer solutions.