Showing that $m^2-n^2+1$ is a square Prove that if $m,n$ are odd integers such that $m^2-n^2+1$ divides $n^2-1$ then $m^2-n^2+1$ is a square number. 
I know that a solution can be obtained from Vieta jumping, but it seems very different to any Vieta jumping problem I've seen.
To start, I chose $m=2a+1$ and and $n=2b+1$ which yields: $$ 4ka^2+4ka-4kb^2-4kb+k = b^2+b$$
Then suppose that $B$ is a solution, and $B_0$ is another solution. Then using Vieta jumping we get (with a bit of algebra) that $B+B_0 = -1$ and $B_0 = \frac {-k(2a+1)^2}{B(4k+1)}$.
But I'm not sure these final equalities are particularly helpful; I can't find any way to yield more solutions from them.  How can I solve the problem? A solution without Vieta jumping is probably also possible

Under the assumption that the integer ratio is positive:
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LEMMA
Given integers $$  M \geq m > 0, $$ along with positive integers $x,y$ with
$$ x^2 - Mxy + y^2 = m.  $$
Then $m$ is a square.
PROOF.
First note that we cannot have integers $xy < 0$ with $ x^2 - Mxy + y^2 = m,  $ since then
 $ x^2 - Mxy + y^2 \geq 1 + M + 1 = M + 2 > m.$ If we have a solution with $x > 0$ and $xy \leq 0,$ it follows that $y=0.$ 
This is the Vieta jumping part, with some extra care about inequalities.
Case I: We begin with integers $$  y > x > 0 $$ and the stronger $$y > Mx.$$
Then we get a new solution by jumping
$$ (x,y) \mapsto (Mx - y,x). $$
However, the assumption $y > Mx$ means $Mx-y < 0,$ we cannot have a solution with one variable positive and the other negative. This case cannot occur.
Case II.  $y > x > 0$ and $y = Mx.$ But then $x^2 - Mxy + y^2 = x^2 - M^2 x^2 + M^2 x^2 = x^2.$ Therefore $x^2 = m$ which is a square.
Case III.
$$ y > x  $$ and
$$ y < Mx.  $$ We have
$$ x^2 - Mxy + y^2 > 0,  $$
$$ x^2 > Mxy - y^2 = y(Mx - y) > x(Mx-y), $$
$$ x > Mx - y > 0. $$
That is, the jump
$$ (x,y) \mapsto (Mx - y,x) $$
takes us from one ordered solution to another ordered solution while strictly decreasing $x+y.$
Within a finite number of such jumps we violate the conditions we were preserving; we reach a solution $(x,y)$ with $y \geq Mx,$ that is
$x > 0$ but $Mx-y \leq 0.$ Since $(Mx - y,x) $ is another solution we know that $Mx-y = 0.$ Therefore $x^2 = m$ and $m$ is a square.
Graph for $x^2 - 5xy + y^2 = 3$

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Suppose we have odd integers $m,n > 0$ such that
$$  \frac{n^2 - 1}{m^2 - n^2 + 1} = k > 0 $$
is an integer. Then 
$$ k+1 = \frac{m^2 }{m^2 - n^2 + 1}.  $$
Name $w = 1 + k,$ so
$$ w = \frac{m^2 }{m^2 - n^2 + 1}.  $$
We are sticking with positive $w$ so we may take $m \geq n >0.$ When we write
$$ m-n = 2x,  $$
$$ m+n = 2y,  $$
we are introducing positive variables. Then $m=x+y,$ $n = y - x,$ and
$$ w = \frac{x^2 + 2xy + y^2}{4xy+1}, $$
$$ x^2 + 2xy+ y^2 = 4wxy + w,  $$
$$ x^2 - (4w-2)xy + y^2 = w. $$
From the LEMMA, we find that $w$ is a square. From
$$ w = \frac{m^2 }{m^2 - n^2 + 1}  $$
we see that 
$$ m^2 - n^2 + 1  $$ is also a square.
