Proof: not a perfect square

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 $(2y-1)^2=4$.

odd squares are $1 \pmod 4,$ but it is more specific than that. Odd squares are $1 \pmod 8.$ You can check this by squaring, say, $1,3,5,7$ and find the remainder when divided by $8$. In particular, squares are never $5 \pmod 8.$ Your $(2y-1)^2 - 4 \equiv 5 \pmod 8$ and cannot be a square