Oh, I think the answer is definitely yes!<br><br>
<br>
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 <i>Glasnik Matematicki</i> we call these solutions
<i>exceptional</i> solutions. Andrej's conjecture is that for any $k$ there is at most 1 exceptional solution.
<br><br>One interesting result we obtain is that, if $k \in K$, then
$y < (2 - \sqrt{3})k$.
<br><br>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 $\{y \to x,k\}$.
<br><br>
It follows that if $k \neq y \pm{1}$, then either $k \in K$ or $y \in K$.
<br><br>
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. $y_n \in K$.
<br><br>
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 $12 \in K$.
<br><br>
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 $k = 1$.
<br><br>
One property shared by all $y_n \in K_1$ is that either $y_n | (x_n+y_n)$ or
$y_n |(x_n-y_n)$.
<br><br>
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 $y_n \in K$.
<br><br>
And of course we can apply the same process to any of these new $y_n$
<i>ad infinitum</i>, 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 $28928, 112 \in K$.
<br><br>
We call these "Type 2" solutions, so let's define $K_2$ to be all of
the $y_n$ found this way. In the paper we show that all 
exceptional solutions can be enumerated recursively in this
fashion, ie. that $K = K_1 \cup K_2$. The enumeration
algorithm is given below. The 3 standard solution classes
are referred to as $0, -1, +1$.<br><br>

<b>Proc <i>Enum_K:</i></b><br><blockquote>
Enum_K1(1,0)<br><br>
for k = 2 to $ \infty $<br><blockquote>
Enum_K1(k, 0)<br>
Enum_K1(k, +1)<br>
Enum_K1(k, -1)</blockquote></blockquote>
<br><br>
<br><br>
<b>Proc <i>Enum_K1(k, class):</i></b><br><blockquote>
set $(x_0, y_0), (x_1, y_1)$ according to class<br>
n1 = if (class = -1 or k = 1) then 2 else 1<br>
for n = n1 to $\inf$<br><blockquote>
add $y_n$ to $K_1$<br>
Enum_K2($y_n$, +1)<br>
Enum_K2($y_n$, -1)</blockquote></blockquote>
<br><br>
<br><br>
Proc <i>Enum_K2</i> is the same as <i>Enum_K1</i> except that it
adds each $y_n$ to $K_2$. Now if Andrej's conjecture is true, which
we believe it most certainly is, then each operation "<i>add</i> $y_n$"
always adds a new $y_n$ to its list, and the two lists $K_1, K_2$ have
no common elements.