Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$ 
Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$. What are the possible values of $$\frac{x^2+y^2-1}{xy}$$?

I have discovered infinitely many pairs. They are $(n, n+1)$. This solution works for all $n$. For all of them the value of $$\frac{x^2+y^2-1}{xy}=2$$
But I want to know if there are any other such solutions. Also what are all the possible values. I have discovered 2. There may be many such values. Are they finite? Can we prove their finite or infinite existence? 
Any help would be appreciated.

Note that the usual problem, and quite difficult if you do not know what to expect, is slightly different, https://en.wikipedia.org/wiki/Vieta_jumping#Example_2
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Define integer $k \geq 3$ as
$$k = \frac{x^2+y^2-1}{xy}$$
We are solving
$$ x^2 - kxy + y^2 = 1. $$
The discriminant of the quadratic form is $\Delta = k^2 - 4$ which is positive but not a square. As a result, if we can solve $\tau^2 - \Delta \sigma^2$ we can construct the generator of the (oriented) automorphism group of the form. Now, $k^2 - \Delta = 4,$ so $\tau = \pm k, \sigma = \pm 1.$
Given a form (positive nonsquare discriminant) of coefficients $\langle A,B,C\rangle$ we can take the generator (in $SL_2 \mathbb Z$) as
$$
\left(
\begin{array}{rr}
\frac{\tau - B \sigma}{2} && -C \sigma \\
A \sigma && \frac{\tau + B \sigma}{2}
\end{array}
\right)
$$
With $\langle A,B,C\rangle = \langle 1,-k,1\rangle$ we get
$$  U = 
\left(
\begin{array}{rr}
k && -1 \\
1 && 0
\end{array}
\right)
$$
and will stick with this one, as convenient for keeping $x \geq y \geq 0.$
The transformation from one solution $(x,y)$ to the next is matrix multiplication with the column vector $(x,y)^T$ on the right. That is,
$$ \color{blue}{ (x,y) \mapsto (kx-y, x)}. $$
The first few solutions with $k \geq 3$ are
$$ (1,0), $$
$$ (k,1), $$
$$ (k^2 - 1,k), $$
$$ (k^3 - 2k,k^2 - 1), $$
$$ (k^4 - 3 k^2 + 1,k^3 - 2k). $$
Oh, as far as actual division, $(1,0)$ is not legal in the original fraction. Life is like that. 
Let's try $k=3$ in the fraction:
$(3,1)$ gives $9/3 = 3.$ $(8,3)$ gives $72/24 = 3.$ 
The final thing is Cayley-Hamilton. The matrix I named $U$ above satisfies $U^2 - k U + I = 0,$ or $U^2 = kU - I.$ As a result, both $x$ and $y$ satisfy linear recurrences,
$$ x_{j+2} = k x_{j+1} - x_j, $$ 
$$ y_{j+2} = k y_{j+1} - y_j. $$
Again with $k=3, $we get $x$ in
$$ 1, 3, 8, 21, 55, 144,  $$
Note that  these are every second Fibonacci number.
With $k=4, $we get $x$ in
$$ 1, 4, 15, 56, 209, 780,  $$
With $k=5, $we get $x$ in
$$ 1, 5, 24, 115, 551, 2640,  $$
With your $k=2$ you have $x^2 - 2xy + y^2 = 1,$ or $(x-y)^2 = 1,$ or $x-y = \pm 1.$
