Find all integer solutions: $x^4+x^3+x^2+x=y^2$ Find all integer solutions of the following equation:
$$x^4+x^3+x^2+x=y^2$$

The equation 
$$y^2 = x^4 + x^3 + x^2 + x = x(x+1)(x^2+1)\tag{*1}$$
has four trivial solutions over $\mathbb{Z}\times\mathbb{Z}$:
$$(x,y) = (-1,0), (0, 0),(1,\pm 2)$$
To determine whether there are other non-trivial solutions, let's look at an equivalent problem:
$$y^2 = \begin{cases}
z(z + 1)(z^2 + 1), &\text{ for } z = x \ge 2\\
z(z - 1)(z^2 + 1), &\text{ for } z = -x \ge 2\tag{*2}
\end{cases}$$
Notice 
$$\gcd(z(z\pm 1),z^2+1) = \gcd( \pm z - 1, z^2 + 1) = \gcd(\pm z - 1,2) = 1 \text{ or } 2$$
Any solution of $(*2)$ must have one of the following two factorizations:
$$\begin{cases}
z(z\pm 1) &= p^2,\\
z^2 + 1   &= q^2,\\
y &= \pm pq;\end{cases}
\quad\text{ or }\quad
\begin{cases}
z(z\pm 1) &= 2p^2,\\
z^2 + 1   &= 2q^2,\\
y &= \pm 2pq.\end{cases}\quad\text{ with } \gcd(p,q) = 1.$$
We can rule out the factorization on the left because for $z \ge 2$, $z^2 + 1$ is never a square. For the factorization on the right, we can rewrite the $x$ part as
$$\left\{\begin{array}{ccrl}
(2 z \pm 1 )^2 &-& 8p^2 &= 1\\
z^2 &-& 2q^2 &= -1
\end{array}\right.\tag{*3}$$
Define $A, B : \mathbb{Z} \to \mathbb{Z}$ by
$$A(k) + B(k) \sqrt{2} = (1+\sqrt{2})^k$$
It is known the positive integer solutions of Diophantine equations like those in $(*3)$
has the form
$$\left\{\begin{array}{ccl}
2 z\pm 1 &=& A(2m)\\
2p &=& B(2m)\\
z &=& A(2n-1)\\
q &=& B(2n-1)\\
\end{array}\right.
\quad\quad\text{ for } m , n \in \mathbb{Z}_{+}.
$$
It is clear $2 z \pm 1 \ge z \implies 2m \ge 2n-1$.
Notice $\frac{A(k+1)}{A(k)} \to 1+\sqrt{2} > 2$ as $k \to \infty$.
For any $n$ large enough such that $A(2n) > 2 A(2n-1) + 1$, there are no way to find
a $m$ to satisfy $(*3)$. A brute force computation shows that this
happens as early as $n \ge 2$. 
This leaves us the cases $n = 1$. It is easy to check
for $n = 1$, one can choose $m = 1$ to make $(*3)$ works. However,
$$
m = n = 1 \quad\implies\quad
\left\{\begin{array}{ccl}
2 z\pm 1 &=& A(2) = 3 \\
2p &=& B(2) = 2\\
z &=& A(1)  = 1\\
q &=& B(1)  = 1\\
\end{array}\right.
\quad\iff\quad
(x,y) = (1,\pm 2)
$$
corresponds to a pair of trivial solutions we have covered before. The conclusion is aside from the four trivial solutions, $(*1)$ doesn't have any non-trivial solution.
