$m$ and $n$ are odd numbers such that $$\large |m^2 - n^2 + 1| \mid (n^2 - 1)$$. Prove that $|m^2 - n^2 + 1|$ is a square number.
I have provided the solution below, but I am uncertain about the structure of my solution, it's kind of clunky in my opinion.
Let $p = \dfrac{n^2 - 1}{m^2 - n^2 + 1} + 1 = \dfrac{m^2}{m^2 - n^2 + 1} \implies (m^2 - n^2 + 1)(p - 1) = (n^2 - 1)$
We need to prove that $p$ is a perfect square.
Let $S$ be a set of integers pairs $(x, y)$ in that order satisfying $$(4xy + 1)p = (x + y)^2 \iff \frac{m^2}{m^2 - n^2 + 1} = \frac{(x + y)^2}{4xy + 1}$$
Let $x = \dfrac{m + n}{2}$ and $y = \dfrac{m - n}{2}$, we have that $$\frac{(x + y)^2}{4xy + 1} = \frac{\left(\dfrac{m + n}{2} + \dfrac{m - n}{2}\right)^2}{4 \cdot \dfrac{m + n}{2} \cdot \dfrac{m - n}{2} + 1} = \frac{m^2}{m^2 - n^2 + 1}$$
Therefore, $\left(\dfrac{m + n}{2}, \dfrac{m - n}{2}\right) \in S \implies S \ne \varnothing$. Let $$a = \min\{|x| \colon \exists(x, y) \in S\}$$
Pointing out $a = 0$ would lead to $p = y^2$ and $p$ is a perfect square.
If $(x, y) \in S$ then $(-x, -y) \in S$.
consequently, there exists $(a, y) \in S$ such that $$(4ay + 1)p = (a + y)^2 \implies y^2 - (4ap - 2a)y + (a^2 - p) = 0$$
Roots $b_1$ and $b_2$ of the equation are integers such that $|b_1| \ge a$ and $|b_2| \ge a$.
(because $(a, b) \in S$, it is also known that $(b, a) \in S$)
On the other hand, we have that $b_1 + b_2 = 4ap - 2a$ and $b_1b_2 = a^2 - p$
We consider two cases.
Case 1: $p < 0$
We have that $b_1 + b_2 \le 0$ and $b_1b_2 > 0 \implies b_1, b_2 < 0$
Because $p < 0$, $a = 0$.
Case 2: $p > 0$ and $a > 0$
We have that $b_1 + b_2 > 0$ and $(a + b_1)(a + b_2) > 0 \implies b_1, b_2 > 0$
Through the selection of $a$, $a^2 \le b_1b_2$, which is contradictory with the equation $b_1b_2 = a^2 - p$.