Diophantine problem: Find all pairs of $(x, y)$ such that $x^3-4xy+y^3=-1$ 
Find all pairs of integers $x, y$ such that:
$$x^3-4xy+y^3=-1$$

My analysis:

*

*$x$ and $y$ can't be both negative because all terms would be negative and $(x,y)=(-1,-1) \implies (-1)^3-4(-1)(-1)+(-1)^3 =-6 $ would already be too small.


*Reducing modulo $4$ gives: $x^3+y^3\equiv 3  \pmod 4
\implies (x,y)\equiv(0,3)$ or $(3,0) \pmod 4$


*WLOG let $x$ be the larger (+ve) one in a solution pair$\implies x\ge y+1$:
$-1 = x^3-4xy+y^3 \ge (y+1)^3 - 4(y+1)y +y^3$
$-2 \ge 2y^3 - y^2 -y$
$-2 \ge (y-1)(2y^2 + y)$
Since $(2y^2+y)\ge 0$ and $y\le-1$
Indeed $(x,y)=(0,-1)$ is a solution fitting all the above (inconclusive) results. However, I don't know how to find a lower bound for $y$ to lock down all possible pairs of $(x,y).$ What's the missing step? Am I completely off?

Hint: A messy quadratic is better than a simple cubic while dealing with integers.
$$x+y=s\implies x^3-4x(s-x)+(s-x)^3=-1 $$

Simplifying, $$ (3s+4)x^2-(3s^2+4s)x+s^3+1=0$$


 The discriminant must be non-negative, $$\Delta=-(3s+4)(s^3-4s^2+4)\ge 0$$


 If $s\in (-\infty,-2]$ or $s\in [4,\infty)$, $\Delta<0$. Therefore, $s\in\{-1,2,3\}$. Only $s=-1$ results in an integer value for $x$.


 $$(x,y)\in \{(-1,0),(0,-1)\}$$

