Show that $m+3$ and $m^2 + 3m +3$ cannot both be perfect cubes.
I've done so much algebra on this, but no luck. Tried multiplying, factoring, etc.
If they were, $$(m+3)(m^2+3m+3)=(m+2)^3+1$$
would be a perfet cube. There aren't many perfect cubes that differ by $1$:
$n^3-(m+2)^3=1$ means $m+2-n\mid1$ and hence $n=m+2\pm1$, reducing the equation to two quadratics: $3(m+2)^2+3(m+2)=0$ and $-3(m+2)^2+3(m+2)-2=0$. This shows that the only value where the statement fails is $m=-2$.
The first observation is that $m=-2$ is a counterexample. Presumably $m$ should be non-negative.
Assume that $m+3=n^3$ is a perfect cube. Then $n>1$ and $$ m^2+3m+3=n^6-3n^3+3. $$ This is strictly less than $n^6=(n^2)^3$. The previous cube below this is $$ (n^2-1)^3=n^6-3n^4+3n^2-1, $$ which is $<n^6-3n^3+3$ as $n\ge2$. Therefore $m^2+3m+3$ is strictly between two consecutive perfect cubes, and cannot be a perfect cube itself.
The statement is false: for $m=-2$, we have $m+3=m^2+3m+3=1$.