6. Find all integer pairs $(x, y)$ such that $x^{3}=y^{3}+2 y^{2}+1$.

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6. Notice that, when $y>0$ or $y<-3$, we always have $y^{3}<y^{3}+2 y+1<(y+1)^{3}$. In this case, $y^{3}+2 y^{2}+1$ is not a cube number, so the original equation has no solution. Therefore, we only need to consider the cases where $y=-3,-2,-1,0$. Substituting these values respectively, we get the integer solutions of the equation as $(x, y)=$ $(-2,-3),(1,-2)$ or $(1,0)$.