It is clear that $(0, \pm 1)$ is a trivial solution.

Multiplying both sides with $4$ gives $$(2y)^2 = (2x^2+x+1)^2 - (x^2-2x-3)$$

When $x^2-2x-3=0$, the above equation yields the following solutions: $(-1, \pm 1)$, $(3,\pm 11)$.

Given there cannot exist another perfect between the square of two consecutive numbers, we must have:

• If $x^2 - 2x-3 >0$, then $(2x^2+x+1)^2 - (x^2-2x-3) \le (2x^2+x)^2$, or
• If $x^2 - 2x-3 <0$, then $(2x^2+x+1)^2 - (x^2-2x-3) \ge (2x^2+x + 2)^2$, or

The first assumption does not hold for any $x$, and the second one will leads to $x=0$ which is already covered.

Therefore, all the solutions are $(-1, \pm1), (0, \pm 1), (3, \pm 11)$