Note that if $ (x,y)$ is a solution in inetegers to $ x^3 - 3xy^2 + y^3 = n$ then $ (y - x, - x)$ and $ ( - y,x - y)$ are also solutions in integers to $ 3 - 3xy^2 + y^3 = n$, because: $ (y - x)^3 - 3(y - x)( - x)^2 + ( - x)^3 = y^3 - 3y^2x + 3x^y - x^3 - 3(x^2y + x^3) - x^3 = x^3 - 3xy^2 + y^3 = n$; $ ( - y)^3 - 3( - y)(x - y)^2 + (x - y)^3 = - y^3 + 3(x^2y - 6y^2x + y^3) + x^3 - 3x^2y + 3y^2x - y^3 = x^3 - 3xy^2 + y^3 = n$. Now, we can never have $ (x,y) = (y - x, - x)$ because this yields $ x = y = 0$ and then $ n = 0$ which is impossible. The same happens when $ (x,y) = (y,x - y)$ and $ (y - x, - x) = (y,x - y)$. Therefore, if $ (x,y)$ is an integer solution to $ x^3 - 3xy^2 + y^3 = n$ then $ (y - x, - x)$ and $ (-y,x - y)$ are two other different solutions in integers to $ 3 - 3xy^2 + y^3 = n$, this proves part one. Assume there do exist integers $ x,y$ such that $ x^3 - 3xy^2 + y^3 = 2891$. Now, $ 2891\equiv 2\pmod 9$; because any cube of an integer is $ - 1,0,$ or $ 1\pmod 9$ then if $ x\equiv 0\pmod 3$ or $ y\equiv 0\pmod 3$ then $ x^3 - 3xy^2\equiv 0\pmod 9$ or $ - 3xy^2 + y^3\equiv 0\pmod 9$, respectively, then $ y^3\equiv 2\pmod 9$ or $ x^3\equiv 2\pmod 9$, respectively which is impossible. Therefore, we cannot have $ x$ or $ y$ congruent to $ 0\pmod 3$. Therefore, $ x^3,y^3$ can be conruent to $ - 1$ or $ 1\pmod 9$. Now, if $ x^3\equiv - 1\pmod 9, y^3\equiv 1\pmod 9$ or $ y^3\equiv - 1\pmod 9, x^3\equiv 1\pmod 9$ then we must have $ - 3xy^2\equiv 2\pmod 9$ which is impossible as $ 3|( - 3xy^2)$. Therefore, we must have $ x^3\equiv y^3\pmod 9$ then $ x^3 + y^3\equiv 2$ or $ - 2\pmod 9$ then $ - 3xy^2\equiv 0$ or $ 4\pmod 9$. However, $ - 3xy^2$ cannot be $ 4\pmod 9$ as $ 3|( - 3xy^2)$. On the other hand, $ - 3xy^2$ cannot be $ 0\pmod 9$ because then $ 3|(xy^2)$ so $ x$ or $ y$ is congruent $ 0\pmod 3$ which we proved before cannot be possible. Therefore, we get a contradiction and the equation $ x^3 - 3xy^2 + y^3 = 2891$ has no integer solutions QED.

Let $\omega$ be a ninth root of unity. Note that $x^3-3xy^2+y^3=(x-y(\omega+\frac{1}{\omega}))(x-y(\omega^2+\frac{1}{\omega^2}))(x-y(\omega^4+\frac{1}{\omega^4}))$. Now note that $(x-y(\omega^2+\frac{1}{\omega^2}))=(x+2y-(\omega+\frac{1}{\omega})^2y)$. Now note that $(\omega+\frac{1}{\omega})(\omega^2+\frac{1}{\omega^2})(\omega^4+\frac{1}{\omega^4})=-1$ and $(\omega+\frac{1}{\omega})(x-y(\omega^2+\frac{1}{\omega^2}))=(-y+(\omega+\frac{1}{\omega})(x-y))$. So using $x^3-3xy^2+y^3=(x-y(\omega+\frac{1}{\omega}))(x-y(\omega^2+\frac{1}{\omega^2}))(x-y(\omega^4+\frac{1}{\omega^4}))$ we get that if $f(x,y)=x^3-3xy^2+y^3$ then $f(x,y)=f(-y,x-y)$ and also $f(x,y)=f(-y,x-y)=f(y-x,-x)$. Hence one solution induces at least three solutions. Now $2891=7^2*59$. Using the $x=t+\frac{1}{t}$ substitution which makes $x^3-3x+1=t^3+1+\frac{1}{t^3}$, then cyclotomic polynomials and field theory implies that for any prime divisor $p$ of $x^3-3x+1$, we have $p$ is $9k \pm 1$ for some $k$. Thus if $7|x^3-3xy^2+y^3$, we see that $7|x,y$. Thus, a solution to $x^3-3xy^2+y^3=2891$ does not exist.