(15) (50 points) Find all positive integers $x, y$ that satisfy the following conditions: (1) $x$ and $y-1$ are coprime; (2) $x^{2}-x+1=y^{3}$.
15 Clearly, $x=1, y=1$ satisfies the requirement.
For $x>1, y>1$, the equation can be transformed into
$$
(y-1)\left(y^{2}+y+1\right)=x(x-1) .
$$

Clearly, $x>y$. Since $(x, y-1)=1$, $x$ must be a divisor of $y^{2}+y+1$. Let $y^{2}+y+1=k x$ ($k$ is a positive integer), thus $x-1=k(y-1)$. From $x>y$, we know $k \geqslant 2$. Eliminating $x$, we get
$$
y^{2}+y+1=k^{2}(y-1)+k,
$$

which is
$$
\left(y^{2}-1\right)+(y-1)=k^{2}(y-1)+k-3 .
$$

This implies $y-1 \mid(k-3)$.
If $k>3$, then $y-1 \leqslant k-3$, i.e., $k \geqslant y+2$, thus
$$
k^{2}(y-1)+k=y^{2}+y+1<k^{2}+(k-2)+1,
$$

which leads to a contradiction $y-1=0$. Therefore, $k \leqslant 3$, so $k=2,3$. Verification shows $y=7$, $x=19$.
In summary, $(x, y)=(1,1),(19,7)$.