Find the real ordered pairs $(x,y)$ satisfying $x^2+x=y^4+y^3+y^2+y.$ Find the real ordered pairs $(x,y)$ satisfying $x^2+x=y^4+y^3+y^2+y.$

$x^2+x=y^4+y^3+y^2+y$
$4x^2+4x=4y^4+4y^3+4y^2+4y$
$4x^2+4x+1=4y^4+4y^3+4y^2+4y+1$
$(2x+1)^2=4y^4+4y^3+4y^2+4y+1$
I am stuck here.Is there a general method to solve such type of equations?Please help me.

HINT: $$x^2+x=y^4+y^3+y^2+y \Rightarrow \left(x+\frac{1}{2}\right)^2=y^3(y+1)+\left(y+\frac{1}{2}\right)^2$$
$$\Rightarrow \left(x+\frac{1}{2}\right)^2-\left(y+\frac{1}{2}\right)^2=y^3(y+1)$$
$$\Rightarrow (x+y+1)(x-y)=y^3(y+1)$$
Now consider the $4$ cases where $x$ is even or odd and $y$ is even or odd and use divisibility of $2$.
Can you proceed now?
