We have given an integer $n \geq 2$ and a positive divisor $m$ of $n$ s.t. $(1) \:\; n = p^3 + m^3$, where $d$ is the smallest divisor of $n$ greater than 1. Clearly $d$ is the smallest prime divisor of $n$. Hence $d=p$, where $p$ is a prime. Hence by equation (1) $(2) \;\; n = p^3 + m^3$, yielding $p \mid m^3$ (since $p \mid n$) and $m \mid p^3$ (since $m \mid n$). Hence $m=p^k$, where $1 \leq k \leq 3$, which inserted in equation (2) give us $(3) \;\; n = p^3 + p^{3k}$. If $p$ is odd, then $n$ is even by equation (3), implying $p=2$. This contradiction means $p$ is even, i.e. $p=2$, which inserted in equation (3) result in $(4) \;\; n = 8 + 8^k$. Consequently, since $k \in \{1,2,3\}$, there are only three integers $n$ satisfying the Diophantine equation (1) under the given conditions, namely $n = 16,72,520$.

A bit quicker: Clearly $d$ is the smallest prime factor of $n$. If $d \ge 3$, then all of $d,m,n$ are odd and we get a contradiction. Hence $d=2$ and the equation is $n=m^3+8$. So $m \mid 8$ and $m$ is even and so $m \in \{2,4,8\}$ which leads to the three solutions $n \in \{16,72,520\}$.

Let $p=d$, $d$ is clearly prime so $p$ is prime. Assume exists prime $q \neq p$ dividing $m$, then $q|d^3$ so $q|p^3$ which is obviously false. So $m$ is a power of $p$. Now if $p=2$, $n=p^3+p^3,n=p^3+p^6,n=p^3+p^9$ all work, i.e. $n=16,72,520$. Otherwise $p>2$ and $p$ is odd, however $n$ is even, so $2|n$, and hence $p=2$, contradiction.