<s>[INST] Find postive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square 
Find positive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square.

My try: Let $\dfrac{x^6-1}{x-1}=y^2$, so $(x^4+x^2+1)(x+1)=y^2$.
Let $d= \gcd(x+1,x^4+x^2+1)$.
And
$d \vert x^4+x^2+1=(x^2+x+1)(x^2-x+1)$, because $d\vert x+1$ so $d\vert 2$.
Case 1: $d=1$. If $d=1$ so $(x^4+x^2+1)$,$(x+1)$ is perfect square and $(x^2+\dfrac{1}{2})^2\le x^4+x^2+1\le(x^2+1)^2$ so no satisfactory solution.
Case 2: $d=2$. If $d=2$ so $\dfrac{x^4+x^2+1}{2}$,$\dfrac{x+1}{2}$ is perfect square. 
But now I stuck. Please give me a hint. Thank you.
 [/INST] You've got the right idea, but as Kenta S's comment indicates, it's $d \mid 3$ (since $x \equiv -1 \pmod{x+1}$ means $x^4 + x^2 + 1 \equiv (-1)^4 + (-1)^2 + 1 \equiv 3 \pmod{x + 1}$), not $d \mid 2$, so the other option is $d = 3$. For that case, for some integers $m$ and $n$, we have
$$x^4 + x^2 + 1 = 3m^2, \; \; x + 1 = 3n^2 \; \; \to \; \; x \equiv -1 \pmod{3} \tag{1}\label{eq1A}$$
With your factorization of
$$x^4+x^2+1=(x^2+x+1)(x^2-x+1) \tag{2}\label{eq2A}$$
note that
$$\begin{equation}\begin{aligned}
\gcd(x^2+x+1,x^2-x+1) & = \gcd(x^2+x+1,x^2+x+1-(x^2-x+1)) \\
& = \gcd(x^2+x+1,2x)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Since $x^2 + x + 1 = x(x+1) + 1$ is always odd, and $\gcd(x^2+x+1,x) = 1$, this means that $x^2+x+1$ and $x^2-x+1$ are always relatively prime. With \eqref{eq1A} giving that $x \equiv -1 \pmod{3}$, then $x^2 - x + 1 \equiv (-1)^2 - (-1) + 1 \equiv 3 \equiv 0 \pmod{3}$. Thus, from the first part of \eqref{eq1A} and using \eqref{eq2A}, there are integers $s$ and $t$ where $m = st$ and
$$x^2 + x + 1 = s^2, \; \; x^2 - x + 1 = 3t^2 \tag{4}\label{eq4A}$$
However, $x^2 \lt x^2 + x + 1 \lt (x + 1)^2$, so the first part of \eqref{eq4A} is not possible. This means there are no integers $x \gt 1$ where $\frac{x^6 - 1}{x - 1}$ is a perfect square.
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