The smallest divisor of $n$ greater than $1$ is the smallest prime divisor of $n$, hence $d$ is prime. Moreover, we have $d \mid n$, hence $d \mid d^3 + m^3$, and $d \mid m^3$. This yields that $m > 1$. On the other hand we have $m \mid n$, hence $m \mid d^3 + m^3$, and $m \mid d^3$. Because $d$ is prime and $m > 1$, we also see that $m$ equals $d$, $d^2$, or $d^3$.

In all cases the parity of $m^3$ is equal to that of $d^3$, and $n = d^3 + m^3$ is even. This means that the smallest divisor of $n$ greater than $1$ equals $2$, i.e. $d = 2$. In case $m = d$, we find $n = 2^3 + 2^3 = 16$, in case $m = d^2$, we find $n = 2^3 + 2^6 = 72$, and in case $m = d^3$, we find $n = 2^3 + 2^9 = 520$. These are indeed solutions: they are even so that $d = 2$, and $2 \mid 16$; $4 \mid 72$ and $8 \mid 520$, which indeed gives $m \mid n$. $\square$