part a) Let $d$ be an arbitrary positive integer. If $k=d(d^2m^6+3dm^3+3)$, then $1+km^3=(dm^3+1)^3$ and $(dm^2n)^3<1+kn^3<(dm^2n+1)^3$ for all positive integers $n<m$. This prove that there exists infinitely many positive integers $k$ which satisfy the condition. part b) Assume that $k$ satisfies the conditon. We have $1+km^3=a^3$ for some positive integer $a$. Since $ p\equiv 2\pmod 3 $, we can say $m^3|a-1$. So we can take positive integer $d$ such that $a=dm^3+1$. So we have $k=d(d^2m^6+3dm^3+3)$. Conversely, if $k=d(d^2m^6+3dm^3+3)$ for some positive integer $d$, then $k$ satisfies the condition as mentioned above. Hence we are done.

kaede wrote: part a) Let $d$ be an arbitrary positive integer. If $k=d(d^2m^6+3dm^3+3)$, then $1+km^3=(dm^3+1)^3$ and $(dm^2n)^3<1+kn^3<(dm^2n+1)^3$ for all positive integers $n<m$. This prove that there exists infinitely many positive integers $k$ which satisfy the condition. How did you find the number $d(d^2m^6+3dm^3+3)$?

To address hangsonkyung's question, let me propose a decoding of what kaede did. First, suppose you couldn't realize first part, and start with part $b)$. The key is that since $p\equiv 2 \pmod{3}$, $-3$ is not a quadratic residue modulo $p$. We will use this as follows: Since $1+kp^{3r}=a^3$ for some $a$, notice that $p^{3r}\mid a^3-1 = (a-1)(a^2+a+1)$. Now, had $a^2+a+1$ been divisible by $p$, we would have had, $$ p\mid a^2+a+1 \implies p\mid (2a+1)^2 + 3 \implies\left(\frac{-3}{p}\right)=1, $$where $(a/p)$ is the Legendre symbol, hence a contradiction. Thus, $a=tp^{3r}+1$ for some $t$, namely, $a=tm^3+1$ for some $t$. Now, work backwards. get $k=t(t^2m^6 + 3tm^3+3)$. This is how you have the example that he (kaede) found.

Notice that if $a^3\equiv 1\pmod{p}$, then $\text{ord}_p(a)\mid 3$ (multiplicative order). If $\text{ord}_p(a)=3$, then by Fermat's Little theorem $a^{p-1}\equiv 1\pmod{p}$, so $\text{ord}_p(a)=3\mid p-1$, contradiction because $p\equiv 2\pmod{3}$, so $\text{ord}_p(a)=1$, so $a\equiv 1\pmod{p}$. Also, you need to mention that $p$ is odd in your statement that if $p\equiv 2\pmod{3}$, then $-3$ is not a quadratic residue mod $p$. See Wikipedia about Quadratic Reciprocity for more information.

Come on, of course, $p$ being odd is just the trivial case $p=2$. (and in this case, $1+km^3=a^3$ is odd, hence, $a^2+a+1$ is automatically not divisible by any power of $2$). What you have done, $a^3 \equiv 1\pmod{p}$ approach, is, almost, precisely how you prove that if $-3$ is a quadratic residue, than $p-1$ must be divisible by $3$.