Suppose that $q=\frac{2^n+1}{3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$ I am curious for a starting point to prove the following claim 

Suppose that $q={2^n+1 \above 1.5pt 3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$

For example take $n=61$. Then $q ={2^{61}+1 \above 1.5pt 3} = 768614336404564651$  which is prime. Then $\binom{2^{61}}{2}-1 = 2658455991569831744654692615953842175$ and it's largest prime factor is $q$. Note ${2^{29}+1 \above 1.5pt 3} =178956971$ is not prime and the largest prime factor of $\binom{2^{29}}{2}-1$ is $3033169$. But even then we could note that $178956971 = 59*3033169$. A counterexample to the claim would  suffice to to show it's wrong. 

First of all, we have to show that $\frac{2^n+1}{3}$ is a factor of $$\binom{2^n}{2}-1=\frac{2^n(2^n-1)}{2}-1=\frac{2^{2n}-2^n-2}{2}$$
To show this, we consider
$$(2^n+1)^2-3(2^n+1)=2^{2n}+2\cdot 2^n+1-3\cdot2^n-3=2^{2n}-2^n-2$$
So, $\binom{2^n}{2}-1$ is divisble by $2^n+1$ and therefore by $\frac{2^n+1}{3}$.
But $(\frac{2^n+1}{3})^2>\frac{2^{2n}+2^n+1}{9}$ 
Since $9$ is a factor of $\binom{2^n}{2}-1$ for odd $n$ (and $n$ must be odd, otherwise $\frac{2^n+1}{3}$ is not an integer) , there cannot be a larger prime factor $p$ because we would have $$3^2\cdot \frac{2^n+1}{3}\cdot p>2^{2n}+2^n+1>\frac{2^{2n}-2^n-2}{2}$$
To show that $9$ is a factor of $2^{2n}-2^n-2$ (for odd n), which implies that $9$ is a factor of $\binom{2^n}{2}-1$ , we consider that $2^n\equiv 2,5 \ or\ 8\mod 9$. In any case we get $2^{2n}-2^n\equiv 2\mod 9$.
