Diophantine equation $n^2+n+1=m^3$

Question

Is there an elementary method for solving Diophantine equation $n^2+n+1=m^3$ for integers $m$ and $n$? There is a similar one, which I could solve:$$p^2-p+1=q^3,$$where $p$ and $q$ are prime numbers. But the technique that I used for solving this doesn't work for the original problem!

Answer 1

$$n^2+n+1=m^3$$ Here can be multiplied by $64$ and obtain the equation: $$(8n-4)^2+48=(4m)^3.$$ All solutions $y^2+48=x^3$ in integers known: $(4;\pm4)$ and $(28,\pm148)$.

Then $m^2-m+1=n^2$ has solutions in positive integers: $(1,1)$ and $(19,7)$.