2.6. (GDR, 74). Solve the equation

$$
(x+2)^{4}-x^{4}=y^{3}
$$

in integers.
2.6. Let the pair of numbers $x, y \in \mathbf{Z}$ satisfy the equation. Suppose that $x \geqslant 0$, then

$$
y^{3}=8 x^{3}+24 x^{2}+32 x+16=8\left(x^{3}+3 x^{2}+4 x+2\right)
$$

Therefore, $y=2 z(2 \in \mathbb{Z})$ and

$$
2^{3}=x^{3}+3 x^{2}+4 x+2
$$

Notice that

$$
(x+1)^{3}=x^{3}+3 x^{2}+3 x+1<z^{3}<x^{3}+6 x^{2}+12 x+8=(x+2)^{3},
$$

so $x+1<z<x+2$, which is impossible. Suppose that $x \leq-2$, then the pair of numbers $x_{i}=-x-2 \geqslant 0, y_{1}=-y$ also satisfies the original equation, since

$$
\left(x_{i}+2\right)^{4}-\left(x_{1}\right)^{4}=x^{4}-(x+2)^{4}=-y^{3}=\left(y y^{3}\right)^{3}
$$

But, as shown above, the inequality $x_{i} \geqslant 0$ leads to a contradiction. Thus, we have the estimates $-2<x<0$, from which we obtain the unique solution $x=-1, y=0$.