Question

I will illustrate the enumeration process with some examples in order to make clear the structure described above.

We start with $k = 1$, the only case with a single solution class $(k, 0)$. We have $k^2+1 = 2$ and $k^2 = 1$. Here is a partial enumeration of all solutions to $x^2 - 2y^2 = 1$:

       n            x            y
       0            1            0
       1            3            2
       2           17           12
       3           99           70
       4          577          408
       5         3363         2378

Because of the symmetry of the equation wrt $k$ and $y$ we know that each pair $(x_n, y_n)$ for $n > 1$ means that $\{y_n \to x_n, 1\}$ is an exceptional solution. For example, we can see that $17^2 - (12^2 + 1).1^2 = 12^2$. Clearly $1 < 12-1$, and so $k = 12$ has an exceptional solution, and because it came from an enumeration of a root class $(1, 0)$ it is a type-1 solution and so we add $12$ to the set $K_1$.

For all $k > 1$ we have 3 root classes, $(k, 0)$, $(k^2-k+1, k-1)$ and $(k^2-k+1, -k+1)$. Partial enumerations for $k=2$ are shown below:

       n            x            y
       0            2            0
       1           18            8
       2          322          144
       3         5778         2584
       4       103682        46368
       5      1860498       832040


       n            x            y
       0            3            1
       1           47           21
       2          843          377
       3        15127         6765
       4       271443       121393

       n            x            y
       0            3           -1
       1            7            3
       2          123           55
       3         2207          987
       4        39603        17711
       5       710647       317811

Each $y_n$ where $n>0$ (or $n>1$ for the 3rd class) provides an exceptional solution $\{y_n \to x_n, 2\}$, and so each $y_n$ is added to $K_1$.

Now every value we add to $K_1$ is an exceptional solution of the form $\{k \to x,y\}$, so for each $k$ in $K_1$ we have an additional pair of conjugate solution classes $(x, \pm{y})$. Assuming the Dujella conjecture is true, these will always be the 4th and 5th classes.

We simply enumerate these classes in similar fashion, except we add the new $y_n$ values to the list $K_2$, since they come from these additional classes for $k \in K_1$, not from the 3 root classes. For example taking exceptional solution $\{18 \to 8,2\}$, we enumerate the classes $(8, 2)$ and $(8, -2)$ for $k=18$:

       n            x            y
       0            18           2
       1          4402         546
       2       1135698      140866
       3     293005682    36342882


       n            x            y
       0            18          -2
       1           242          30
       2         62418        7742
       3      16103602     1997406

Again, every $(x_n, y_n)$ for $n>0$ gives a new exceptional solution $\{y_n \to x_n,18\}$, and so we add each $y_n$ to $K_2$. And every item $k$ we add to $K_2$ represents 2 new classes for that $k$, so we can apply the same procedure recursively to each and every one.

The reason that I have kept $K_1$ and $K_2$ as two distinct lists is that the members of $K_1$ have properties not shared by $K_2$. The divisibility property noted above is one such property, another is the fact that all of the root classes for any $k$, from which we poulate $K_1$, have explicit polynomial descriptions, which lend themselves to the sort of analysis that we can't readily apply to $K_2$.

For example, we can (I believe) deduce from the properties of these polynomials that every operation "add $y_n$ to $K_1$" provides a unique value. I also believe that I will be able It remains to demonstrate that no value in $K_2$ be seen whether we can also occur in $K_1$. If this is in fact true, then prove the only remaining hurdle is a proof that every operation "add $y_n$ to same holds for $K_2$" adds a unique value.K_2$.