= Solution
Use the final printed restriction that $m$ is even. Square-freeness then gives $m\equiv2\pmod4$, and the integer ring is $\mathcal O_K=\mathbb Z[\sqrt{-m}]$. An integer solution has $x>0$. If $y$ were even, $y^2+m$ would have 2-adic valuation one, impossible for a cube; hence $x,y$ are odd. For an odd prime $p\mid m$, $p\mid y$ would similarly give valuation one. Thus $\gcd(y,m)=1$.
The ideals generated by $y+\sqrt{-m}$ and $y-\sqrt{-m}$ are coprime. Any common prime ideal divides their sum and difference and therefore lies above a rational prime dividing $2m$ and $y$. The odd possibilities have just been excluded; above two, $\sqrt{-m}$ is zero modulo the ramified prime while $y$ is odd, so neither factor is divisible by it. <Unique factorization of ideals in a number field> in
$$
(y+\sqrt{-m})(y-\sqrt{-m})=(x)^3
$$
now yields $(y+\sqrt{-m})=\mathfrak a^3$. Since the <class number> is coprime to three, the ideal class of $\mathfrak a$ is trivial. The ideal is principal, and the only units for this imaginary quadratic field with $m\geq5$ are $\pm1$, which can be absorbed into a cube. Hence
$$
y+\sqrt{-m}=(u+v\sqrt{-m})^3,\qquad u,v\in\mathbb Z.
$$
Comparing coefficients of $\sqrt{-m}$ gives
$$
v(3u^2-mv^2)=1.
$$
Thus $v=1$ with $3u^2=m+1$, or $v=-1$ with $3u^2=m-1$. Both alternatives cannot occur, since $m-1$ and $m+1$ cannot both be divisible by three. The eligible alternative permits at most the two choices $u=\pm\sqrt{(m\pm1)/3}$, and then
$$
x=u^2+m,\qquad y=u(u^2-3m).
$$
Consequently \b[there are at most two integer solutions $(x,y)$], which, when present, have the same $x$ and opposite $y$. The proof uses the evenness assumption; it does not silently claim the same coprimality argument for every odd $m$.
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