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 to demonstrate that no value in $K_2$ can also occur in $K_1$. If this is in fact true, then the only remaining hurdle is a proof that every operation "add $y_n$ to $K_2$" adds a unique value.