Suppose that \(x,y,z\) are nonzero elements of
\(\mathbb{Z}[\omega]\) that satisfy \((\ast)\). We can assume w.l.o.g.
that \(x,y,z\) are pairwise coprime.

Consider the number \(\rho=1-\omega\). It is prime, as its norm equals
\((1-\omega)(1-\omega^2)=3\). We observe that
\(\overline{\rho}=1-\omega^2=(1-\omega)(1+\omega)\sim\rho\); hence
\(\alpha\in\mathbb{Z}[\omega]\) is divisible by \(\rho\) if and only if
so is \(\overline{alpha}\). Each element in \(\mathbb{Z}[\omega]\) is
congruent to \(-1,0\) or 1 (mod \(\rho\)): indeed, \(a+b\omega\equiv
a+b=3q+r\equiv r\) (mod \(\rho\)) for some \(q\in\mathbb{Z}\) and
\(r\in\{-1,0,1\}\).

The importance of number \(\rho\) lies in the following property:
\[\alpha\equiv\pm1\; (\mbox{mod }\rho)\;
(\alpha\in\mathbb{Z}[\omega])\;\;\mbox{implies}\;\;
\alpha^3\equiv\pm1\;(\mbox{mod }\rho^4). \quad\quad\quad\quad\quad(2)\] Indeed, if
\(\alpha=\pm1+\beta\rho\), we have \(a^3\mp1=(a\mp1)(a\mp\omega)
(a\mp\omega^2)=\rho^3\beta(\beta\pm1)(\beta\pm(1+\omega))\), where
the elements \(b\), \(b\pm1\), \(b\pm(1+\omega)\) leave three distinct
remainders modulo \(\rho\), implying that one of them is also
divisible by \(\rho\), thus justifying our claim.

Among the numbers \(x,y,z\), (exactly) one must be divisible by
\(\rho\): otherwise, by (2), \(x^3,y^3,z^3\) would be congruent to
\(\pm1\) (mod \(\rho^4\)), which would imply one of the false
congruences \(0\equiv\pm1\), \(\pm1\equiv\pm2\) (mod \(\rho^4\)). We
assume w.l.o.g. that \(\rho\mid z\). Moreover, (2) also gives us that
\(\rho^2\mid z\).

Let \(k\geq2\) be the smallest natural number for which there exists a
solution to \((\ast)\) with \((x,y,z)=1\) and \(\rho^k\mid z\),
\(\rho^{k+1}\nmid z\). Consider this solution \((x,y,z)\).

The factors \(x+y\), \(\omega x+\omega^2y\), \(\omega^2x+\omega y\) from
(1) are congruent modulo \(\rho\) and have the sum 0. It follows from
\(\rho\mid z\) that each of them is divisible by \(\rho\) and that
\(\rho\) is their greatest common divisor. Let
\[x+y=A\rho,\;\;\;\omega x+\omega^2y=B\rho,\;\;\;\omega^2x+\omega
y=C\rho,\] where \(A,B,C\in\mathbb{Z}[\omega]\) are pairwise coprime
and \(A+B+C=0\). The product \(ABC\) is a perfect cube (equal to
\((z/\rho)^3\)), and hence each of \(A,B,C\) is adjoint to a cube:
\[A=\alpha\zeta^3,\;\;\;B=\beta\eta^3,\;\;\;C=\gamma\xi^3\] for some
pairwise coprime \(\zeta,\eta,\xi\in\mathbb{Z}[\omega]\) and units
\(\alpha,\beta,\gamma\). Therefore,
\[\alpha\zeta^3+\beta\eta^3 +\gamma\xi^3=0.\quad\quad\quad\quad\quad(3)\] Since
\(\alpha\beta\gamma\) is a unit and a perfect cube, we have
\(\alpha\beta\gamma=\pm1\). Furthermore, \(ABC=(z/\rho)^3\) is divisible
by \(\rho\) (since \(\rho^2\mid z\)), so (exactly) one of the numbers
\(\zeta,\eta,\xi\), say \(\xi\), is divisible by \(\rho\). In fact,
\(\xi^3\) divides \(ABC\) which is divisible by \(\rho^{3k-3}\) and not by
\(\rho^{3k-2}\), so \(\rho^{k-1}\) is the greatest power of \(\rho\) that
divides \(\xi\). The numbers \(\zeta\) and \(\eta\) are not divisible by
\(\rho\); consequently, \(\zeta^3\) and \(\eta^3\) are congruent to \(\pm1\)
modulo \(\rho^4\). Thus the equality \(A+B+C=0\) gives us
\(\alpha\pm\beta\equiv0\) (mod \(\rho^4\)), therefore \(\beta=\pm
\alpha\); now \(\alpha\beta\gamma=\pm1\) implies \(\gamma=\pm\alpha\).

Canceling \(\alpha\) in equation (3) yields
\(\zeta^3\pm\eta^3\pm\xi^3=0\), which gives us another nontrivial
solution to \((\ast)\) with \((\zeta,\eta,\xi)=1\). However, in this
solution we have \(\rho^{k-1}\mid\xi\) and \(\rho^k\nmid\xi\), which
contradicts the choice of \(k\).