Equation (1) is symmetric in $x$ and $y$, so we can initially assume $x \geq y$. For $x \neq y$, for each solution $(x, y)$, $(y, x)$ is also a solution. With
$d=x-y \geq 0$ it follows that $\sqrt[3]{7 d^{2}+x y}=d+1$. Squaring yields $x^{2}-d x+\left(-d^{3}+4 d^{2}-3 d-1\right)=0$ with the discriminant $D=d^{2}-4\left(-d^{3}+4 d^{2}-3 d-1\right)=(d-2)^{2}(4 d+1) \geq 0$. Therefore, $x_{1 / 2}=\frac{d \pm(d-2) \sqrt{4 d+1}}{2}$. For $x$ to be an integer, $4 d+1$ must be a perfect square, and specifically the square of an odd number. The approach $4 d+1=(2 m+1)^{2}$ with $m \in\{0 ; 1 ; 2 ; \ldots\}$ yields $d=m^{2}+m$, so that $x_{1 / 2}=\frac{1}{2}\left[\left(m^{2}+m\right) \pm\left(m^{2}+m-2\right)(2 m+1)\right]$ results. This leads to $\left(x_{1}, y_{1}\right)=\left(m^{3}+2 m^{2}-m-1, m^{3}+m^{2}-2 m-1\right)$ and for $m \neq 1$ to $\left(x_{2}, y_{2}\right)=\left(-m^{3}-m^{2}+2 m+1,-m^{3}-2 m^{2}+m+1\right)$ and the solutions $\left(y_{1}, x_{1}\right)$ and $\left(y_{2}, x_{2}\right)$ for $m>0$. A check confirms that these pairs are indeed solutions.

Hint: The simple special cases were found by most participants. Often, attempts were then made (in vain) to show that there are no further solutions using estimates.