Let $k>m>0,$ $k,m \in \mathbb{N},$ such that $$k^3-m^3 \mid km(k^2-m^2)$$ Determine $k,m.$
It's clear that $$(k-m)(k^2+km+m^2) \mid km(k-m)(k+m)$$ . So $k^2+km+m^2 \mid km(k+m)$ But $gcd(k^2+km+m^2,km(k+m)=1$ Does this imply that such numbers only exist if $k^2+km+m^2=1$? But then there wouldn't exist such numbers because k>m>0.
You've made a a good start but, as Thomas' comment states, you're assuming that $\gcd(k, m) = 1$, which the problem doesn't state is a condition. Thus, instead have
$$\gcd(k, m) = d, \;\; k = dk_1, \;\; m = dm_1, \;\; \gcd(k_1, m_1) = 1 \tag{1}\label{eq1A}$$
Since the condition $k \gt m \;\to\; k_1 \gt m_1 \;\to\; k_1 - m_1 \gt 0$, we then get
$$\begin{equation}\begin{aligned} d^3(k_1^3 - m_1^3) & \mid d^4(k_{1}m_{1}(k_1^2 - m_1^2) \\ (k_1 - m_1)(k_1^2 + k_{1}m_{1} + m_1^2) & \mid dk_{1}m_{1}(k_1 - m_1)(k_1 + m_1) \\ k_1^2 + k_{1}m_{1} + m_1^2 & \mid dk_{1}m_{1}(k_1 + m_1) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
As you stated, we have
$$\gcd(k_1^2 + k_{1}m_{1} + m_1^2, k_{1}m_{1}(k_1 + m_1)) = 1 \tag{3}\label{eq3A}$$
This is because $k_1^2 + k_{1}m_{1} + m_1^2 \equiv m_1^{2} \pmod{k_1}$, $k_1^2 + k_{1}m_{1} + m_1^2 \equiv k_1^{2} \pmod{m_1}$ and $k_1^2 + k_{1}m_{1} + m_1^2 \equiv m_1^{2} \pmod{k_1 + m_1}$, where the $\gcd(k_1, m_1) = 1$ part in \eqref{eq1A} means that each result is relatively prime to the modulo. Thus, we then must have
$$k_1^2 + k_{1}m_{1} + m_1^2 \mid d \;\;\to\;\; d = n(k_1^2 + k_{1}m_{1} + m_1^2), \;\; n \ge 1 \tag{4}\label{eq4A}$$
Altogether, this results in the general solution of
$$k_1 \gt m_1 \gt 0, \; \gcd(k_1, m_1) = 1, \; n \ge 1, \; d = n(k_1^2 + k_{1}m_{1} + m_1^2), \; k = dk_{1}, m = dm_{1} \tag{5}\label{eq5A}$$
Note that Thomas' comment solution of $k = a(a^2 + ab + b^2)$, $m = b(a^2 + ab + b^2)$ is a special case of \eqref{eq5A}. Also, a simple example is where $m_1 = 2$, $k_1 = 1$, $n = 1$ giving that $d = 7$, $k = 14$ and $m = 7$ where we get that $2401 \mid 14406$ since $14406 = 6\times 2401$.
