Prove that $x^2 + x +1 = y^2$ has no positive integer solutions

Question

I began the proof by factoring the LHS to be $x(x+1)+1 = y^2$. Since the LHS contains the product of two consecutive integers, that part is even. But since $1$ is added, it must be odd. That means $y$ must also be odd. How should I continue from here?

Answer 1

Hint:

If $x^2+x+1=y^2$ with $x$ a positive integer, then $x^2<y^2<(x+1)^2=x^2+2x+1$

Answer 2

$x^2+x+1=y^2$ implies $4x^2+4x+4=(2y)^2$ or $(2y)^2-(2x+1)^2 = 3$. The only integer squares that differ by $3$ are $4$ and $1$, so the only solutions are given by $(x,y)\in\{(0,1),(0,-1),(-1,1),(-1,-1)\}$.