When you pass from $2^k$ to $2^{k+1}$, the number doubles. If you could show that passing from $k^3$ to $(k+1)^3$ increases the number by a factor less than $2$, you’d be in business. How big is $\frac{(k+1)^3}{k^3}$? If $k\ge 10$, then
$\frac{(k+1)^3}{k^3}=\left(\frac{k+1}k\right)^3=\left(1+\frac1k\right)^3\le\left(1+\frac1{10}\right)^3=\left(\frac{11}{10}\right)^3=\frac{1331}{1000}<2\;.\tag{1}$
Thus, if $2^k>k^3$ and $k\ge 10$, we have
$2^{k+1}=2\cdot 2^k>\frac{(k+1)^3}{k^3}\cdot 2^k>\frac{(k+1)^3}{k^3}\cdot k^3=(k+1)^3\;,$
exactly as desired.
Writing it out in one go like that may make it look more mysterious than it really is. All I really did was ask myself what happens to the two sides of the inequality $2^k>k^3$ when $k$ is replaced by $k+1$. Clearly the lefthand side is doubled. What happens to the righthand side (in terms of multiplicative increase) isn’t so obvious, but at least we can say that it gets multiplied by $\frac{(k+1)^3}{k^3}$:
$\begin{array}{c} &2^k&>&k^3\\ \text{multiply by }2&\downarrow&&\downarrow&\text{multiply by }\frac{(k+1)^3}{k^3}\\ &2^{k+1}&\overset{?}>&(k+1)^3 \end{array}$
Clearly the inequality in the bottom row will be true if $2\ge\frac{(k+1)^3}{k^3}$, so we just have to make sure that this is the case $-$ which is exactly what I did with the calculation $(1)$.