This is problem from Russian mathematical olympiad 1997. It can be found in book MATHEMATICAL OLYMPIADS FROM AROUND A WORLD 1997-1998 ublished my MAA It is problem 32 Russia 1977 page 97 or 98. Here is link so you can download this book. http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf

After looking at the elegant proof in the link above, I decided to destroy my rough sketch...

(m^2-n^2)^2=16n+1. 16n+1=(m^2-n^2)^2>=0.So, 16n+1>=0,but as n is integer, 16n+1=0 doesn't take place.So,16n+1>0 and because of this n>=1. Let abs(m)>=abs(n)+1.Then, (m^2-n^2)^2-16n+1>=((n+1)^2-n^2)^2= =(2n-1)^2. So, 16n+1>=(2n-1)^2=4n^2-4n+1, or 4n^2-20n=4n(n-5)<=0. As n>0,n-5<=0.So,n=1,n=2,n=3,n=4 or n=5. By checking we have that the solutions are pairs (1;0),(-1;0),(4;3),(-4;3),(4;5),(-4;5). If abs(m)<=abs(n)-1, we will have the same as in the previous: y<=5. answer:(1;0),(-1;0),(4;3),(-4;3),(4;5),(-4;5).

My solution: If $|m|>n$ $\Longrightarrow$ $16n+1=(m^2-n^2)^2\geq (2n+1)^2$ $\Longrightarrow$ $16n+1\geq 4n^2+4n+1$ $\Longrightarrow$ $3\geq n$ $\Longrightarrow$ if $n=0$ then $m=1,-1$ $\Longrightarrow$ $(m,n)=(1,0);(-1,0)$ are solutions if $n=1,2$ it has no solutions since $17,33$ are not perfect squares if $n=3$ then $m=4,-4$ $\Longrightarrow$ $(m,n)=(4,3);(-4,3)$ are solutions If $n>|m|$ $\Longrightarrow$ $16n+1=(m^2-n^2)^2\geq (2n-1)^2$ $\Longrightarrow$ $16n+1\geq 4n^2-4n+1$ $\Longrightarrow$ $5\geq n$ If $n=1,2,4$ it has no solutions since $17,33,65$ are no perfect squares If $n=5$ then $m=4,-4$ $\Longrightarrow$ $(m,n)=(4,5);(-4,5)$ If $n=0,3$ solutions have been found in the previous case Then all solutions are: $(m,n)=(4,3);(-4,3);(4,5);(-4,5);(1,0);(-1,0)$...