Seeking examples of proofs [by contradiction?] using Pell equation I’m looking for proofs which use Pell equations as a critical part of the proof. One example is this paper by Robert Phillips: on page 4, he says

If $(p, q)$ is one of the infinite number of solutions to $x^2−ky^2 = 1$, then…

and leverages that fact to complete his proof.
What other [elementary] proofs do something similar? How many fundamentally different ways can the Pell equation be leveraged in a proof?
In particular, I’m interested in proofs by contradiction, i.e., proofs which say something like

“Assume $x$ is not a square, so that there exist integers $p$ and $q$ such that $p^2-xq^2=1$…”

and then continue on to prove [by way of contradiction] that $x$ actually is a square.
 A: Pell's equation can be used to prove by contradiction that $\sqrt 2$ is irrational. See, for example, here. The idea used in the answer can be extended to prove that if $N$ is a nonsquare positive integer, then $\sqrt N$ is irrational.
Claim 1 : If $N$ is a nonsquare positive integer, then $\sqrt N$ is irrational.
Proof : Assume $\sqrt{N}$ is rational. Then, we can write $\sqrt N= \dfrac{s}{t}$ where $s,t$ are positive integers. Consider the iterations $x_{n+1} = \dfrac{x_n}{2}+\dfrac{N}{2x_n}$ which is the formula given by Newton's method for the function $f(x) = x^2 -N$ with a special initial value $x_0 = \dfrac{U}{V}$ where $(U,V)$ is the fundamental solution of the Pell's equation $u^2-Nv^2=1$.
Let $x_n = \dfrac{p_n}{q_n}$. Then, we get
$$p_{n+1} = p_n^2 + Nq_n^2,\qquad q_{n+1} = 2p_n q_n,$$
$$p_0^2-Nq_0^2 = U^2-NV^2 = 1,\qquad p_{n+1}^2-Nq_{n+1}^2 = (p_n^2 -Nq_n^2)^2=1.$$
So, $(p_n,q_n)$ is a solution to the Pell's equation $u^2-Nv^2=1$.
From the above, we have $$\dfrac{N}{x_n}< \sqrt{N} < x_n\tag1$$ and we can get
\begin{equation}
\bigg|\dfrac{p_n}{q_n}-\dfrac{s}{t}\bigg| = |x_n - \sqrt{N}| < \bigg|x_n - \dfrac{N}{x_n}\bigg| =\bigg|\dfrac{p_n^2-Nq_n^2}{p_nq_n}\bigg|= \dfrac{1}{p_n q_n}.
\end{equation}
So $|tp_n-sq_n| < \dfrac{t}{p_n}$. Since $p_n$ grows to infinity, we have $|tp_n - sq_n|<1$ for some $n$. Then $tp_n- sq_n=0$ implies $x_n =\sqrt N$ which contradicts $(1)$.$\quad\blacksquare$

Pell's equation also can be used to prove by contradiction that an equation has no solution. For example, this answer proves by contradiction that the equation $\dfrac{m}{n}+\dfrac{n+1}{m}=4$ has no solution in positive integers.  The idea used in the answer can be extended to prove the following claim. The case in the answer is the case when $(a,b,c,d)=(6,1,2,2)$ in the following claim.
Claim 2 : Given positive integers $a\ (\ge 3),b,c,d$ such that the fundamental solution $(u_1,v_1)$ of the Pell's equation $u^2-adv^2=1$ satisfies $u_1\equiv 1\pmod a$, the equation $\dfrac{b^2dm}{n}+\dfrac{(c^2d-a)n+2}{m}=2bcd$ has no solution in positive integers.
Proof : Assume the equation has a solution in positive integers.
Multiplying the both sides of the equation by $-amn$ gives
$$-ab^2dm^2-a(c^2d-a)n^2-2an+2abcdmn=0,$$
i.e.
$$a^2n^2-2an-ad(b^2m^2-2bcmn+c^2n^2)=0$$
which can be written as
$$(an-1)^2-ad(bm-cn)^2=1.$$
If $bm-cn=0$, then $an-1=\pm 1$ which is impossible. So, $bm-cn\not=0$. Letting $u=an-1$ and $v=|bm-cn|$, we have $$u^2-adv^2=1.\tag2$$
If $ad=s^2$ where $s$ is an integer, then $(u-sv)(u+sv)=1$ implies $u=an-1=\pm 1$ which is impossible. So, $ad$ is not a square. Then, the general solution in positive integers $\text{mod}\ a$ of the Pell's equation $(2)$ can be written as
$$\pmatrix{u\\v}\equiv \pmatrix{1&0\\v_1&1}^k\pmatrix{1\\v_1}\pmod a.$$
Since we obtain $$\pmatrix{1&0\\x&1} \pmatrix{1&0\\v_1&1}=\pmatrix{1&0\\x+v_1&1}$$it follows that $u\equiv 1\pmod a$ which contradicts $u=an-1$.$\quad\blacksquare$
