2. Find all pairs of positive integers $(x, y)$ such that
$$
\sqrt[3]{7 x^{2}-13 x y+7 y^{2}}=|x-y|+1 .
$$
2. $(x, y)=(1,1)$ or
$$
\{x, y\}=\left\{m^{3}+m^{2}-2 m-1, m^{3}+2 m^{2}-m-1\right\},
$$

where, integer $m \geqslant 2$.
Let $(x, y)$ be any pair of positive integers satisfying equation (1).
If $x=y$, then
$$
x^{\frac{2}{3}}=1 \Rightarrow x=1 \Rightarrow(x, y)=(1,1) \text {. }
$$

If $x \neq y$, by symmetry, we can assume $x>y$.
Let $n=x-y$. Then $n$ is a positive integer.
Thus, equation (1) can be rewritten as
$$
\sqrt[3]{7(y+n)^{2}-13(y+n) y+7 y^{2}}=n+1 \text {. }
$$

Cubing both sides of the above equation and simplifying, we get
$$
\begin{array}{l}
y^{2}+y n=n^{3}-4 n^{2}+3 n+1 \\
\Rightarrow(2 y+n)^{2}=4 n^{3}-15 n^{2}+12 n+4 \\
\quad=(n-2)^{2}(4 n+1) .
\end{array}
$$

When $n=1,2$, there are no positive integer solutions for $y$ that satisfy equation (2).

When $n>2$, $4 n+1$ is the square of the rational number $\frac{2 y+n}{n-2}$, so it must be a perfect square.

Since $4 n+1$ is odd, there exists a non-negative integer $m$ such that
$$
\begin{array}{l}
4 n+1=(2 m+1)^{2} \\
\Rightarrow n=m^{2}+m .
\end{array}
$$

By $n>2$, we know $m \geqslant 2$.
Substituting equation (3) into equation (2), we get
$$
\begin{array}{l}
\left(2 y+m^{2}+m\right)^{2}=\left(m^{2}+m-2\right)^{2}(2 m+1)^{2} \\
=\left(2 m^{3}+3 m^{2}-3 m-2\right)^{2} . \\
\text { Also, } 2 m^{3}+3 m^{2}-3 m-2 \\
=(m-1)\left(2 m^{2}+5 m+2\right)>0,
\end{array}
$$

then $2 y+m^{2}+m=2 m^{3}+3 m^{2}-3 m-2$.
Thus, $y=m^{3}+m^{2}-2 m-1$.
By $m \geqslant 2$, we know
$$
y=\left(m^{3}-1\right)+(m-2) m>0,
$$

and
$$
\begin{array}{l}
x=y+n=m^{3}+m^{2}-2 m-1+m^{2}+m \\
=m^{3}+2 m^{2}-m-1
\end{array}
$$

is also a positive integer.