a family of Pellian equations
This is an interesting question. Let me explain why I believe that the answer is no.
Rewrite the equation in the form $$ (x-k)(x+k) = (k^2+1)y^2. $$ Interpreting this as an equation in the polynomial ring ${\mathbb Z}[k]$ and using unique factorization you end up with equations of the form $$ x - k = (k^2+1)A^2, \quad x + k = B^2, $$ which implies $$ 2k = B^2 - (k^2+1)A^2. $$ The solution $A = 1$ and $B = k+1$ gives rise to one of yours.
In order to be able to find a lot more fundamental solutions you will have to substitute $k = f(t)$ for a polynomial $f$ that makes $k^2+1$ reducible. the simplest choice is $k = 2t^2$ since $$ k^2+1 = 4t^4 + 1 = (2t^2+1)^2 - 4t^2 = (2t^2+2t+1)(2t^2-2t+1). $$ Going through the game above will reveal your additional solution in this case.
If you substitute $ k = t + 2t^3$ (I hope I remember my calculations well), the polynomial $k^2+1$ splits into three quadratic factors. Whether the resulting Pell type equations have more solutions is difficult to check. But my impression is that of you can find many substitutions for which $k^2+1$ splits into many factors, chances are that you get more than 2*2+1 = 5 fundamental solutions. Where's Noam when you need him?
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.