Oh, I think the answer is definitely yes!

Let $\{k \to x,y\}$ be any solution of $x^2 - (k^2+1)y^2 = k^2$, and let $K$ be the set of $k$ for which a solution has $0 < k < y-1$. In a paper recently
submitted to *Glasnik Matematicki* we call these solutions
*exceptional* solutions. Andrej's conjecture is that for any $k$ there is at most 1 exceptional solution.

One interesting result we obtain is that, if $k \in K$, then
$y < (2 - \sqrt{3})k$.

A particular feature of this Pell equation is its symmetry wrt $k$ and $y$. These are interchangeable, so for any solution $\{k \to x,y\}$ there is a corresponding solution $\{y \to x,k\}$.

It follows that if $k \neq y \pm{1}$, then either $k \in K$ or $y \in K$.

Now, for any $k \geq 2$, we have 3 particular solution classes $(x_n, y_n)$ with
$y_0 = \{0, k-1, -(k-1)\}$. For any $n > \{0, 0, 1\}$ we have $y_n > k-1$
and so $\{y_n \to x_n,k\}$ is exceptional, ie. $y_n \in K$.

We also need to consider $k=1$, for which there is just the one
class with $(x_0, y_0) = (1, 0)$ , $(x_1, y_1) = (3, 2)$ and so for
any $n > 1$ we have $y_n \in K$. For example $(x_2, y_2) = (17, 12)$ from
which we obtain exceptional solution $\{12 \to 17, 1\}$, and so $12 \in K$.

In our paper we call the corresponding set of exceptional solutions
"Type 1". But here let us simply define the set $K_1$ to be all of these
$y_n > k$ that we find from these 3 classes for any $k > 1$, and
from the one class for $k = 1$.

One property shared by all type 1 solutions, ie $\{k \to x, y\}$ with $k \in K_1$ is that either $(y^2 +1) | (x+y)$ or $(y^2 + 1) |(x-y)$.

Now, for any $k \in K_1$ we have a corresponding $\{k \to x,y\}$ for
which our Pell eqn has 2 additional solution classes, with fundamental solutions
$(x_0, y_0) = (x, \pm y)$. For any $n > \{0,1\}$ we then have $y_n > k-1$
and so $\{y_n \to x_n, k\}$ is exceptional, ie $y_n \in K$.

And of course we can apply the same process to any of these new $y_n$
*ad infinitum*, each $y_n$ seeding a forest of others. For example,
just considering $n = 1$ alone in each case, from $\{8 \to 18,\pm{2}\}, 8 \in K_1$
we obtain $\{546 \to 4402,8\}$ and $\{30 \to 242,8\}$, so $546, 30 \in K$,
and from $\{30 \to 242, \pm{8}\}$ we get $\{28928 \to 868322,242\}$ and
$\{112 \to 3362,30\}$, so $28928, 112 \in K$.

We call these "Type 2" solutions, so let's define $K_2$ to be all of
the $y_n$ found this way. These do not have the divisibility property that
was noted above for the $y_n \in K_1$.

In the paper we show that all
exceptional solutions can be enumerated recursively in this
fashion, ie. that $K = K_1 \cup K_2$. This is done by showing any $k \in K$ can
be traced back to a root in $K_1$.

The enumeration
algorithm is given below. Solution classes are referred to as $0, -1, +1$, the interpretation of which I hope is reasonably clear!

**Proc ***Enum_K:*

Enum_K1(1,0)

for k = 2 to $ \infty $

Enum_K1(k, 0)

Enum_K1(k, +1)

Enum_K1(k, -1)

**Proc ***Enum_K1(k, class):*
set $(x_0, y_0), (x_1, y_1)$ according to class

n1 = if (class = -1 or k = 1) then 2 else 1

for n = n1 to $\infty$

add $y_n$ to $K_1$

Enum_K2($y_n$, +1)

Enum_K2($y_n$, -1)

**Proc ***Enum_K2(k, class):*
set $(x_0, y_0), (x_1, y_1)$ according to class

n1 = if (class = -1) then 2 else 1

for n = n1 to $\infty$

add $y_n$ to $K_2$

Enum_K2($y_n$, +1)

Enum_K2($y_n$, -1)

To generate the solution sequences in any class, we note that each class has the same
recurrence relation:

$R = 2k^2 + 1$

$x_n = 2Rx_{n-1} - x_{n-2}$

$y_n = 2Ry_{n-1} - y_{n-2}$

but of course have different initial conditions:

$R = 2k^2 + 1, S = 2k, D = k^2 + 1$

$K_1, class 0: (x_0, y_0) = (k, 0), (x_1, y_1) = (kR, kS)$

$K_1, class +: (x_0, y_0) = (k^2-k+1, k-1)$

$K_1, class -: (x_0, y_0) = (k^2-k+1,1-k)$

$K_2, class +: (x_0, y_0) = (x_n, +y_n)$ for any $y_n \in K_1 \cup K_2$

$K_2, class -: (x_0, y_0) = (x_n, -y_n)$ " "

and in all cases above $(x_1, y_1)$ satisfy

$x_1 = Rx_0 + DSy_0$

$y_1 = Ry_0 + Sx_0$

Now if Andrej's conjecture is true, and we believe it is, then each operation "

*add* $y_n$" always adds a new $y_n$ to its list, and the two lists $K_1, K_2$ have
no common elements.

An implementation of

*Enum_K* with a bailout parameter finds
that with $k <10^6$ we have $|K_1| = 882, |K_2| = 163, |K| = 1045$, and
that every $k$ enumerated was unique. This agrees with Andrej's figure.