Diophantine equation of 2nd order

Question

So I have a test in arithmetics tomorrow and was exercising a bit, until I came across this problem:

Solve: $$x^2 = 4y^2 +3$$ where $x,y\in\mathbb{Z}$.

My knowledge in this field is very limited (I'm only aware of Bezout's, theorem fo GCDs, Gauss's theorem ($a\mid bc$ and $a∧b=1 \iff a|c$), and Fermat's little theorem, so I'd appreciate clarification).

Answer 1

Hint: $$x^2-4y^2=(x-2y)(x+2y)=3$$

Can you continue from here?

Answer 2

The equation $x^2 = 4 z+3$ has no solutions in $\mathbb{Z}$.