Consider the even polynomial, $$P_m(x)=\prod_{i=1}^{2m}\left(1+x^{2 i^3}\right).$$ I want to prove that for any $N$, there is some value of $m$ for which the expansion of $P_m$ contains a sequence of $N$ non-zero terms with even powers.
For example, if $N=4$, then $m=7$ suffices since the expansion of $P_7$ has $4$ consecutive terms with even powers: $$x^{110},x^{112},x^{114},x^{116}.$$ Numerically, and intuitively, this seems to be well-supported, but I don't know where to begin a proof, or if one can even be found. As much as I'd love this result, it does feel a little like the subset sum problem, which is known to be NP-complete.
First substitute $y=x^2$. We are now looking for consecutive integer powers of $y$ that are included in the expansion of $$\prod_{i=1}^{2m} \left(1+y^{i^3}\right)$$ The term $y^n$ is included in this expansion iff it can be expressed as the sum of distinct cubes in the set $\{1,8,27,\ldots 8m^3\}$.
We can choose a sufficiently large $m$, and we are now looking for consecutive integers that can be expressed in the form $$\sum_{n=1}^\infty e_n n^3~\text{ where }~e_n\in\{0,1\}$$
According to oeis, the set of numbers that can be expressed as the sum of distinct cubes is a finite set whose largest element is $12758$.
Hence, for sufficiently large $m$ (a very loose lower bound could be $2m>\sqrt[3]{12758+N}$), the expansion of the generating function will contain at least the nonzero terms with consecutive powers ranging from $12759$ to $12758+N$.