For such equations:

$$\frac{x^2+y^2}{xy+1}=t^2$$

Using the solutions of the Pell equation. $p^2-(t^4-4)s^2=1$

You can write the solution.

$$x=4tps$$

$$y=t(p^2+2t^2ps+(t^4-4)s^2)$$

It all comes down to the Pell equation - as I said.

Considering specifically the equation:

$$\frac{x^2+y^2}{xy-1}=5$$

Decisions are determined such consistency. Where the next value is determined using the previous one.

$$p_2=55p_1+252s_1$$

$$s_2=12p_1+55s_1$$

You start with numbers. $(p_1;s_1) - (55 ; 12)$

Using these numbers, the solution can be written according to a formula.

$$y=p^2+2ps+21s^2$$

$$x=3p^2+26ps+63s^2$$

If you use an initial $(p_1 ; s_1) - (1 ; 1)$

Then the solutions are and are determined by formula.

$$y=s$$

$$x=\frac{p+5s}{2}$$

As the sequence it is possible to write endlessly. Then the solutions of the equation, too, can be infinite.

If you use a sequence with the first element. $ (p ; s ) $ - $( 4 ; 1 )$
Then decisions can be recorded.

$$y=2s$$

$$x=p+5s$$

If you use a sequence with the first element. $( p ; s )$ - $( 55 ; 12 )$

Using this sequence can be different. On its basis with the first element.
$( z ; q )$ - $(2 ; 1 )$

$$z_2=pz_1+7sq_1$$

$$q_2=pq_1+3sz_1$$

Then decisions will be.

$$x=z-q$$

$$y=z+q$$

It is necessary to take into account that the number can have a different sign. - $(p ; s )$