$m^2-n^2+1 \mid n^2-1\Rightarrow m^2-n^2+1 \mid m^2$ Substitution $m=x+y,n=x-y$ ($m$ and $n$ are integers of like parity) $\frac{(x+y)^2}{4xy+1}=k,\;k\in\mathbb{N}$ Use Vieta jumping to prove that $k$ is a square. Therefore $4xy+1=m^2-n^2+1$ is square too.

$m^2-n^2+1\mid n^2-1$ Let $(m^2-n^2+1)k=n^2-1$ $$k=\frac{n^2-1}{m^2-(n^2-1)}$$$$\Rightarrow k+1=\frac{m^2}{m^2-(n^2-1)}$$$$\Rightarrow m^2-(n^2-1)\mid m^2$$Therefore we have $m^2=(m^2-(n^2-1))(k+1)$ Thus if we prove $k+1$ is a perfect square then the question is done. $$km^2=(n+1)(n-1)(k+1)$$...(1) Now $k , k+1$ are coprime $$\Rightarrow k\mid (n+1)(n-1)$$Thus we have two cases : case 1) $\frac{n^2-1}{k}=k+1$ $$k(k+1)=(n+1)(n-1)$$$$gcd(n+1,n-1)=2 ; gcd(k,k+1)=1$$therefore no solutions. case 2) $\frac{n^2-1}{k}=(k+1)x^2$ $$\Rightarrow n^2-1=k(k+1)x^2$$only solution is $x=2$ which gives $n-1=2k$ or $n=2k+1$ putting in (1) we get : $m^2=4(k+1)^2$ which gives $m , n$ of opposing parity.(contradiction) Therefore $(k+1)$ has to be a perfect square , which implies that $m^2-n^2+1$ is also a perfect square.