If $7$ Divides $\binom{2^n}{2}+1$ then $n =3k+2$ for some positive integer $k$ It is a straightforward question.

If $$ 7 \text{ }\Bigg | \binom{2^n}{2}+1$$ then $n=3k+2$ for some
  positive integer $k$.

This is just curiosity  no motivation just rummaging through some old question in a notebook. A simple counter example would work. 
Note if $\mathfrak a(n) =\binom{2^n}{2}+1$ then $7$ divides $\mathfrak a(n)$ for $n=2,5,8,11,14,17,\ldots$ and from this I am guessing that $n=3k+2$

Something slightly more general: we have ${k\choose 2} \equiv -1 \pmod{7}$ if and only if $k\equiv 4\pmod{7}$.
To see this, write ${k\choose 2} = \frac{k(k-1)}{2}$, so the equation is equivalent to $k^2 - k \equiv -2\pmod{7}$.  But $k^2 - k + 2 \equiv (k-4)^2 \pmod{7}$, so this is satisfied exactly when $k\equiv 4\pmod{7}$.
Finally, it is easy to see that $2^n$ cycles through the values $1,2,4\pmod{7}$, and the result follows.
