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I also believe that I will be abledemonstrate that no value in$K_2$also occur in$K_1$. If this is in fact true, thenonly remaining hurdle is a proof that every operation "add$y_n$ to" adds a unique value

It remainsbe seen whether weprovesame holds for

       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

"$add y_n to K_1$ amconfident"$add y_n to K_2$

       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

 
"add$y_n$ to$K_1$believe"add$y_n$ to$K_2$

       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 $k, y$ 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 $17^2 - (12^2 + 1).1^2 = 12^2$, 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 $n>0$ ($n>1$ for the 3rd class) provides an exceptional solution $\{y_n \to x_n, 2\}$, and each $y_n$ is thus 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 $y_n$ 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

keepyet

       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

recursively and everyhave keptreadily