<s>[INST] Is $x^2+x+1$ ever a perfect power? Using completing the square and factoring method I could show that the equation $x^2+x+1=y^n$, where $x,y$ are positive odd and $n$ is positive even integers, does not have solution, but I could not show that  for positive odd $x,y$ and odd $n>1$ the equation does (does not) have solution.
Thank you for your contribution.
 [/INST] Your equation can be rewritten as
$$\frac{x^{3}-1}{x-1} = y^{N}.$$
The Diophantine equation
$$ \frac{x^{n} − 1}{x-1} = y^{q} \quad x > 1, \quad y>1, \quad n>2 \quad q \geq 2 \quad \tag1$$
was the subject matter of a couple of papers of T. Nagell from the $1920's.$ Some twenty-odd years later, W. Ljunggren clarified some points in Nagell’s arguments and completed the proof of the following result.
Theorem. Apart from the solutions
$$\frac{3^{5}-1}{3-1}=11^{2}, \quad \frac{7^{4}-1}{7-1}=20^{2}, \quad \frac{18^{3}-1}{18-1} = 7^{3},$$
the equation in $(1)$ has no other solution $(x, y, n, q)$ if either one of the following conditions is satisfied:
$(1.$ $q = 2$,
$(2.$ $3$ divides $n$,
$(3.$ $4$ divides $n$,
$(4.$ $q =3$ and $n$ is not congruent with $5$ modulo $6$.
Clearly enough, this theorem implies that there are only two solutions to the equation you are considering: $(x=1, y=3, N=1)$ and $(x=18, y=7, N=3)$.
</s>