<s>[INST] 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 $n=3k+2$
 [/INST] $$\binom{2^n}2+1=\frac{2^n(2^n-1)}2+1=2^{2n-1}-2^{n-1}+1$$
Now, let's apply mod $7$. Note that $2^3\equiv 1\pmod 7$.
If $n=3k$ we have $4-4+1\neq 0$.
If $n=3k+1$ we have $2-1+1\neq 0$.
This proves your statement, but we can check easily if the "iff" statement is true:
If $n=3k+1$ we kave $1-2+1=0$.
So $7$ divides $\binom{2^n}n+1$ if and only if $n\equiv 2\pmod 3$.
</s>