Video solution:

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Written solution:

Suppose $x^3$ is any perfect cube in this range, where $x$ is a positive integer.

If $x^3 \leq 2^{18}+1,$ then $$x^3 = 2^{18}+1$$ or $$x^3 \leq 2^{18} = {2^6}^3$$$$\implies x \leq 2^6 = 64 .$$ If $x^3 = 2^{18}+1 = 64^3 +1,$ then it follows that $$64^3 < x^3 < 65^3$$$$\implies 64 < x < 65$$ This would mean that $x$ isn't an integer. This is a contradiction, so we know $$x \leq 64 .$$ We also know $$2^8+1= 257 \leq x^3 .$$ Now, suppose $257\leq x^3 < 343.$ Then, we know $216 < x^3 < 343 .$

This means $6 < x < 7 ,$ which also means that $x$ isn't an integer. This is a contradiction, so $7\le x.$

Therefore, all $x$ which satisfy $343 \leq x^3 \leq (2^6)^3$ must also satisfy $$7 \leq x \leq 64 .$$ Therefore, the number of possible $x$'s is $64-7+1 = 58.$

Thus, E is the correct answer.