Observe that for non-negative integers $x$ and $y$ that satisfy the equation $x^3 + 7x^2 + 35x + 27 = y^3$, we obtain
$$\begin{aligned} y^3 - (x+2)^3 &= x^3 + 7x^2 + 35x + 27 - x^3 - 6x^2 - 12x - 8 \\ &= x^2 + 23x + 19 > 0 \end{aligned}$$
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.