This problem can be solved using the squeeze method. The given relation can be re-written as: $$(2x^2+x)^2 + (3x^2+4x+4) = 4y^2 = (2x^2+x+1)^2 - (x^2-2x-3)$$ It is impossible for $y^2$to be between to consecutive squares, $(2x^2+x)^2$ and $(2x^2+x+1)^2 $. Therefore $(3x^2 + 4x+ 4)$ and $(x^2-2x-3)$ cannot be both positive. The value of $(3x^2 + 4x+4)$ is always positive because $$3x^2 + 4x+4=2x^2 + (x^2 +4x+4)=2x^2 + (x+2)^2$$ Consequently, the value of $(x^2-2x-3)$ must be negative, i.e. $$x^2-2x-x \le 0 \implies -1 \le x \le 3$$ Setting $x=-1, 0, 1, 2, 3$, respectively leads to the following solutions: $$(x, y) = \boxed{(-1, \pm 1), (0, \pm 1), (3, \pm11)}$$