Diophantine equation with quartic polynomial 
What are all integral solutions to
$$y^2=x^4+x^3+x^2+x+1$$

The RHS could become
$$(x^2-x+1)(x^2+x+1)+x(x^2+1)$$
or
$$\frac{x^5-1}{x-1}$$
I have no idea how to manipulate the equation into something useful or what the first step should be.
Also, A quartic diophantine equation looks useful, but none of the answers completely solve the question? Thanks!
 A: If $x \gt 0$,
the only solution is
$x=3, y=11$.
Note:
Wolfy was used extensively
getting this answer.
$y^2
=x^4+x^3+x^2+x+1
$
$y^2
=x^4+x^3+x^2+x+1
\gt x^4
\implies
y > x^2
$.
$(x+1)^4
=x^4+4x^4+,,,
$
so
$y^2
\lt (x+1)^4
$
or
$y < (x+1)^2
$.
$(x^2+x/2+3/8)^2
=x^4 + x^3 + x^2 + (3 x)/8 + 9/64
\lt y^2
$
so
$y > x^2+x/2+3/8$.
$(x^2+x/2+1)^2
=x^4 + x^3 + (9 x^2)/4 + x + 1
\gt y^2
$
so
$y < x^2+x/2+1
$.
If $x = 2n$ then
$4n^2+n+3/8
\lt y
\lt 4n^2+n+1
$,
so there can be no such
integer $y$.
If $x = 2n+1$ then
$x^2+x/2+3/8
=4n^2+4n+1+n+1/2+3/8
=4n^2+5n+15/8
$
and
$x^2+x/2+1
=4n^2+4n+1+n+1/2+1
=4n^2+5n+5/2
$
so
$y
= 4n^2+5n+2
$.
But
$y^2
=16 n^4 + 40 n^3 + 41 n^2 + 20 n + 4
$
and
$x^4+x^3+x^2+x+1
=\dfrac{x^5-1}{x-1}
=\dfrac{(2n+1)^5-1}{2n}
=16 n^4 + 40 n^3 + 40 n^2 + 20 n + 5
$
and the difference is
$n^2-1$
so they are never equal
unless $n = 1$
so $x = 3
$.
A: There is a result of Masser [1] that $$y^2 = f(x) \tag 1$$ where $f(x)$ is a monic polynomial and not a perfect square, any integer solution $(x,y)$ of Eqn. (1) satisfies the inequality
$$
|x| \le 26 H(f)^3 \tag 2
$$
where $H(f)$ denotes the classical height of $f$ given by the maximum of the absolute values of the coefficients of $f(x)$.
In the given equation, $H(f) = 1$. Therefore, all integer solutions are given by $x \in [-26, +26]$.
This range is quite small and effective for a computer search.
A brute force search in this range yields
$$(x,y) \in \{(-1,1), (0,1), (3,11)\}, \text{ where } (x,y) \in \mathbb{Z\times Z}.$$

Note: Checking for small root $x \in \{0,\pm 1, \pm 2, \cdots \}$ and eliminating a factor to reduce to a cubic or quadratic is the normal method followed. We could have used that as suggested in other answers and comments. The method above is useful when such an elimination is not possible or tedious. Therefore, I am providing this method for reference of other readers who may find this numerical method useful, in general.

References
[1]: Masser, D. W. “Polynomial Bounds for Diophantine Equations.” The American Mathematical Monthly, vol. 93, no. 6, 1986, pp. 486–88. JSTOR, https://doi.org/10.2307/2323487. Accessed 6 Dec. 2022.
