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. 

Most probably yes, because:
$$\binom{2^n}{2}-1=\frac{2^n(2^n-1)}{2}-1=2^{n-1}(2^n-1)-1=$$
$$=2^{n-1}(2^n+1 - 2)-1 = 2^{n-1}(2^n+1) - 2^n-1=(2^n+1)(2^{n-1}-1)=$$
$$=3q(2^{n-1}-1)$$
Additionally:
$$2 \equiv -1 \pmod{3}$$
$$2^n \equiv (-1)^n \pmod{3}$$
$$2^n + 1 \equiv (-1)^n +1 \pmod{3}$$
So $n$ is odd, further leading to
$$2^{n-1}-1=\left(2^{\frac{n-1}{2}}-1\right)\left(2^{\frac{n-1}{2}}+1\right)$$
