How to prove indirectly that there exist no positive integers $x$ and $y$ for which $x^2 - y^2 = 1$?

Question

I already did it with a contraposition. But how to do it indirectly? It has to be something like "there exist positive integers $x$ and $y$ for wich $x^2 - y^2 = 1$".

Tried to use $(x+y)(x-y)=1$ and follow that $x-y=1$ and $x+y=1$ so that $y=-y$, which cannot be true with positive integers. Possible?

Answer 1

Since the difference of two consecutive squares is: $$(n+1)^2-n^2=2n+1,$$ and $2n+1=1\implies n=0$, the result follows.

Answer 2

$$(n+i)^2-n^2=2in+i^2$$

So, for any $n>0$ we have an increasing function of $i$, and vice versa. If $i\geq 1$ then $$(n+i)^2-n^2\geq 2n+1\implies n\leq0$$

But, we are looking for positive numbers $n$. Also, it may be obvious but if $x<y$ then there is no way for $x^2-y^2=1$.

Another way of looking at it would be $$\forall n,i\in\mathbb N \;2in+i^2\geq 3$$