Integer solutions to $x^3=y^3+2y+1$? 
Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$

My approach:
I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.

Solution
Assume that $$x=y+t,~~~t \in \mathbb{Z}.$$
Then $$(y+t)^3=y^3+2y+1,$$ namely $$3ty^2+(3t^2-2)y+t^3-1=0.\tag1$$
If $t=0$, then $y=\dfrac{1}{2}$, which is absurd. Therefore, $t \neq 0$, which implies that $(1)$ could be seen as  a quadratic equation with respect to $y$.
Consider the discriminant for $(1)$. $$\Delta=(3t^2-2)^2-4\cdot 3t \cdot (t^3-1)=-3(t^2+2)^2+12t+16.$$
If $|t|\geq 2$, then $$\Delta=-3(t^2+2)^2+12t+16<-3(t+2)^2+12t+16=4-3t^2<0.$$ Thus, $(1)$ has no real root.
Therefore, the possible values of $t$ are $t=\pm 1.$ Now, we may verify that $t=1$ is the only one solution. Under this case, $$x=1,y=0.$$
