Prove: If n is a perfect square, then n+3 is not a perfect square Use either direct proof, proof by contrapositive, or proof by contradiction.
Using proof by contradiction method

Assume n is a perfect square and n+3 is a perfect square (proof by
contradiction)
There exists integers x and y such that $n = x^2$ and $n+3 = y^2$
Then $x^2 + 3 = y^2$
Then $3 = y^2 - x^2$
Then $3 = (y-x)(y+x)$
Then $y+x = 3$ and $y-x=1$
Then $x = 1, y = 2$
Since $x = 1$, that implies $n = 1$
this is how far I got

Anyone know what I should do now?

First, as mentioned by others, $1$ is a perfect square and $1+3$ is a perfect square. So you need to prove "If $N > 1$ is a perfect square, then $n+3$ is not a perfect square.
Let $n = m^2$ where $m$ and $n$ are positive integers and $n > 1$. Then we must have $m > 1$. Since $m$ is an integer, we can say that $m \ge 2$. Hence $2m+1 \ge 5$. So
$$(m+1)^2 = m^2 + (2m+1) \ge n + 5 > n+3$$
Since $n < n+3 < n+5$, then $m^2 < n+3 < (m+1)^2$
