Why $x^4+x=y^2+y$ has only a finite number of integer solutions? In an attempt to understand and solve this problem, I tried to play with some small finite example, one of which is $$x^4+x=y^2+y$$ Playing with Wolfram-Alpha indicates indeed equations of similar form, where a generic parametric solution (i.e. $x,y$ can both be represented as some function of $t$) does not exist, always have only a finite number of integer solutions.
It seems progress can be made if I am able to understand why there is only a finite number of integer solutions for some specific examples. If anyone has any idea please share with me.
 A: If $x^4+x = y^2 + y$,
Then $4(x^4+x) = 4(y^2 + y)$
or $(2y + 1)^2 = 4x^4+4x + 1$
But you need form $a^2x^2 \pm 2abx + b^2$ for infinite solution and the RHS polynomial cannot be converted to that form.
EDIT: Adding further why $4x^4+4x + 1$ polynomial cannot be a square for integer $x \gt 1$. I have taken only positive integer $x$ for the proof, as an example.
If the polynomial is a square,
$
\begin {align}
4x^4+4x + 1 &= n^2, \space n \in \mathbb{Z+} \\
4x + 1 &= n^2 - 4x^4 \\  
4x + 1 &= (n-2x^2)(n+2x^2) \\
\end {align}
$
As ($4x + 1$) is positive for $x \ge 0$, $n - 2x^2 \gt 0$.
Say, $n = a + 2x^2$ where $a \in \mathbb{Z+}$
$4x + 1 = a (4x^2 + a)$
This holds true for $a = 1$ and $x = (0, 1)$. But for $x \ge 2$, LHS will be smaller than the RHS.
A: for positive, both bigger than $1$
if $y \geq x^2,$ then $y^2 + y \geq x^4 + x^2 > x^4 + x$
if $1 < y \leq x^2 - 1,$ $y^2 + y \leq x^4 - x^2 < x^4 + x$
I guess there are 3 more cases $x < -1, y > 1,$  then $x > 1, y < -1,$ then $x < -1, y < -1$
