Irska

Observe that for non-negative integers $x$ and $y$ that satisfy the equation $x^3+7x^2+35x+27=y^3$, we obtain $$\begin{array}{rl}{y}^3-(x+2{)}^3& =x^3+7x^2+35x+27-x^3-6x^2-12x-8\\ & =x^2+23x+19>0\end{array}$$ and similarly $$(x+4{)}^3-y^3=5x^2+13x+37>0$$ Thus $x+2<y<x+4$, and since $x$ and $y$ are integers, $y=x+3$. Substitution in the original equation yields $$x^3+7x^2+35x+27=x^3+9x^2+27x+27$$ and so $x^2=4x$. Hence $x=0$ or $x=4$, and the corresponding values of $y$ are $3$ and $7$. The pairs $(0,3)$ and $(4,7)$ are easily seen to satisfy the equation.