The answer is $x = y = 1$ and $\{x, y\} = \{m^3 + m^2 - 2m - 1,\ m^3 + 2m^2 - m - 1\}$, where $m \ge 2$.

1. If $x = y$, then the original equation is equivalent to $x^{2/3} = 1$, so $x = y = 1$.

2. If $x > y$, let $n = x - y$, then the original equation can be rewritten as
$$\sqrt[3]{7(y + n)^2 - 13(y + n)y + 7y^2} = n + 1.$$
Cubing both sides and simplifying, we have
$$y^2 + yn = n^3 - 4n^2 + 3n + 1.$$
To complete the square on the left side, we multiply both sides by $4$ and add $n^2$ to both sides, obtaining
$$(2y + n)^2 = (n - 2)^2(4n + 1).$$
Clearly $n \le 2$ is impossible. When $n > 2$, since $4n + 1$ must be a perfect square, we must have $4n + 1 = (2m + 1)^2$, hence
$$n = m^2 + m, \tag{1}$$
where $m \ge 2$ (since $n \ge 3$). Substituting back into the original equation, we get
$$(2y + m^2 + m)^2 = (2m^3 + 3m^2 - 3m - 2)^2.$$
Thus clearly $2y + m^2 + m = 2m^3 + 3m^2 - 3m - 2 \Leftrightarrow y = m^3 + m^2 - 2m - 1$.
Substituting back gives the solution.