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Claim. For a fixed index $k$ (the paper states no range; its derivation and its check at $k=1$ treat $k$ as a positive integer),

$$\sum_{n\ge0}\frac{1}{F_{2^nk}}= \begin{cases} \dfrac{2L_k-F_{2k}\sqrt5+5F_k^2}{2F_{2k}} & k \text{ odd},\\[2ex] \dfrac{2-F_k\sqrt5+L_k}{2F_k} & k \text{ even}, \end{cases}$$

where $L_k$ is the $k$-th Lucas number. In both cases the coefficient of $\sqrt5$ is $-1/2$ and the other terms are rational, so each sum is irrational and the instances $n_j=2^jk$ of Problem 267 have answer yes; that inference is this page's, since the paper evaluates the sums and does not discuss their irrationality. The case $k=1$ recovers Good's value $(7-\sqrt5)/2$. The paper gives no theorem numbers; the closed form is stated on its page 455 (result page). The method telescopes the identity $F_{2k}=F_kL_k$ with the Lucas identities $L_{m+p}+L_{m-p}=L_mL_p$ ($p$ even) and $L_k^2=L_{2k}+2(-1)^k$, sums the resulting Lucas series with a summation formula of Siler, and passes to the limit through powers of $(1\pm\sqrt5)/2$. The source is V. E. Hoggatt, Jr. and Marjorie Bicknell, A reciprocal series of Fibonacci numbers with subscripts $2^nk$, Fibonacci Quart. 14 (1976), no. 5, 453–455, on the card hoggattjr_1976_reciprocal_series_fibonacci_numbers_subscripts.

Covers. The instances $n_j=2^jk$ for every positive integer $k$, the case $k=1$ being Good's instance $n_j=2^j$ (the Good page): the answer is yes. Not covered: every other index sequence. All these instances lie inside Badea's 1993 condition (the Badea page).

Acceptance. Refereed: The Fibonacci Quarterly, volume 14, number 5 (December 1976). The site's commentary credits Bicknell and Hoggatt, with Good, with the irrationality of the sum over the indices $2^n$ but labels the problem OPEN, so that commentary is not listed as reviewed evidence. The corpus has not reproved the closed form and awards no tier of its own.

Depends on. Nothing in this wiki; the claim rests on the cited paper.