We know that IF $n$ is a perfect square, THEN $n=3q$ or $n=3q+1$ for some $q$. This theorem is false the other way around.
What I'm saying is, if $n$ is not a perfect square, then we can still have $n=3q$ or $n=3q+1$.
Basically it comes down to logic. "If $A$, then $B$" is not the same as "If not $A$, then not $B$", but it is the same as "If not $B$, then not $A$". If it rains, I get wet. If I get wet, it doesn't have to rain - I could be in the shower. But if I don't get wet, we know for sure that it doesn't rain.
This is why you can use the theorem
if $n$ is a perfect square, then $n=3q$ or $n=3q+1$ for some $q$.
To say
if not $n=3q$ or $n=3q+1$ for some $q$, then not $n$ is a perfect square.
which in normal words would be
if $n\neq3q$ or $n\neq3q+1$ for all $q$, then $n$ is not a perfect square.
So in your first example, $98765432$ is not of the form $98765432=3q$ and also not of the form $98765432=3q+1$, thus, it cannot be a perfect square.
Best Answer
For the sake of contradiction write $(2y-1)^2-4=n^2$ where $n$ is an integer. Equivalently $$4=(2y-1-n)(2y-1+n).$$ Difference between the two factors is $2n$, i.e. even. Only ways to factor $4$ with factors that differ by even number are $(-2)\cdot(-2)$ and $2 \cdot 2$, both cases are impossible as they imply $n=0$ and $(2y-1)^2=4$.