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It also follows from the generating function for $\infty$-bonacci numbers (here I'm starting with the 0th term equal to 1), $$\frac{1}{1-x-x^2 -x^3 -\cdots}= \frac{1}{1-x/(1-x)}=1+\frac{x}{1-2x}.$$
If we take the limit of the $i$th $n$-bonacci number as $n\to\infty$ we get $2^{i-1}$ and $\prod_{i=2}^\infty (1-x^{2^{i-1}}) = \sum_{i=0}^\infty \pm x^i$.
It might be noted that the identity can be generalized to $$\sum_{i=0}^m \binom{m+a-i}{a}\binom{2m+a+1}{i} = 4^m\binom{m+a/2}{m}.$$ The original identity is the case $m=n-1, a=2$.
The generalization referred to in the remark is $$S(m,n)= \sum_{k=0}^{m-l} (-1)^k 2^{2m-2k-2l}\binom{m-l}{k} S(l, n+k) .$$ Identities (1) and (2) are the cases $l=0$ and $l=1$.
It should be noted that this generating function for $f_N(x)$ (in a slightly different form) is also found in Zagier's paper, Elementary Aspects of the Verlinde Formula and of the Harder-Narasimhan-Atiyah-Bott Formula, Israel Math. Conf. Proc. 9 (1996), archive.mpim-bonn.mpg.de/id/eprint/2138/1/preprint_1994_5.pdf
I added some information to the OEIS page (oeis.org/A247239). In particular there is a generating function $$\sum_{m=1}^\infty (2/N)^m F(m,N) x^{2m}=1-\frac{Nx \cot(N\sin^{-1}x)}{\sqrt{1-x^2}}$$ (references for this formula to papers of Richard Stanley and me are on the OEIS page). This formula makes it easy to compute $F(m,N)$ for any fixed $m$, but I don't see that it leads to a proof of integrality.