Show that for any $n$, there are infinitely many cubes of the form $2^na - 9$.
Progress: We use induction on $n.$ For $n=1$ it works. Say it works for $n-1$. We will show for $n$. Note that $2^na-9$ is $$2^{n-1}2a-9.$$ If we have infinite $a$ such that $a$ is even for $n-1$, then we are done. Else, only finite amounts of $a$ are even.
So after a large constant $N$, we have $2^{n-1}a-9=x^3$ for only odd $a$.
So say $$2^{n-1}a_1-2^{n-1}a_2=u^3-w^3\implies 2^{n-1}|u^3-w^3$$
But I couldn't get anything further.
I thought about taking $v_2$, as in, if $2^na=x^3+9$ then we can take $x=y^2$. So we have $$2^na=y^2+3^2.$$