Is $1+{n\choose 2}+{n\choose 4}$ ever a power of two except for $n=1,2,3,4,5,10$?

Question

In this 3Blue1Brown video the formula $F=1+{n\choose 2}+{n\choose 4}$ is used to count the maximum number of regions that $n$ chords can cut a circle. The famous Moser's circle problem goes as follows: When $n=1,2,3,4,5$ we get that the maximum number of regions that $n$ chords can cut a circle is $1,2,4,8,16$ (respectively), but when $n=6$ we unexpectedly get the number $31$. In the video, Grant explains why this is the case using Pascal's triangle and the formula $F=1+{n\choose 2}+{n\choose 4}$ where $F$ denotes the maximum number of regions that $n$ chords can cut a circle. At the end of the video he conjectures that $1+{n\choose 2}+{n\choose 4}$ is not a power of two for all values of $n\neq 1,2,3,4,5,10$. He said he thinks it can be done using Diophantine equations, but he doesn't know how to do it. It goes with out saying that neither do I.