Let $S = x + y$ and $P = x y$.

We are given:

$$S = x + y = x^2 + y^2 = x^3 + y^3$$

Recall:
$$x^2 + y^2 = (x + y)^2 - 2 x y = S^2 - 2P$$
x^3 + y^3 = (x + y)^3 - 3 x y (x + y) = S^3 - 3 P S
$$$$

So, the system becomes:

1. $S = S^2 - 2P$
2. $S = S^3 - 3 P S$

From (1):
$$S = S^2 - 2P \implies S^2 - S - 2P = 0 \implies 2P = S^2 - S \implies P = \frac{S^2 - S}{2}$$

From (2):
$$S = S^3 - 3 P S \implies S^3 - 3 P S - S = 0 \implies S^3 - S - 3 P S = 0$$
Substitute $P$ from above:
$$S^3 - S - 3 S \cdot \frac{S^2 - S}{2} = 0$$
Multiply both sides by 2:
$$2 S^3 - 2 S - 3 S (S^2 - S) = 0$$
$$2 S^3 - 2 S - 3 S^3 + 3 S^2 = 0$$
$$(2 S^3 - 3 S^3) + (3 S^2) - 2 S = 0$$
$$(-S^3) + 3 S^2 - 2 S = 0$$
$$S^3 - 3 S^2 + 2 S = 0$$
$$S (S^2 - 3 S + 2) = 0$$
$$S (S - 1)(S - 2) = 0$$

So the possible values for $S$ are $0$, $1$, or $2$.

Now, for each $S$, compute $P$:

- If $S = 0$:
$$P = \frac{0^2 - 0}{2} = 0$$
So $x$ and $y$ are roots of $t^2 - 0 t + 0 = 0 \implies t^2 = 0 \implies t = 0$
So $(x, y) = (0, 0)$

- If $S = 1$:
$$P = \frac{1^2 - 1}{2} = 0$$
So $x$ and $y$ are roots of $t^2 - t = 0 \implies t (t - 1) = 0$
So $(x, y) = (0, 1)$ and $(1, 0)$

- If $S = 2$:
$$P = \frac{2^2 - 2}{2} = \frac{4 - 2}{2} = 1$$
So $x$ and $y$ are roots of $t^2 - 2 t + 1 = 0 \implies (t - 1)^2 = 0 \implies t = 1$
So $(x, y) = (1, 1)$

Therefore, all solutions are:
$$(x, y) = (0, 0),\ (0, 1),\ (1, 0),\ (1, 1)$$