<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] $7|\binom{2^n}{2}+1\iff \binom{2^n}{2}\equiv -1 \bmod 7\iff 2^n(2^{n}-1)2^{-1}\equiv -1 \bmod 7\iff $
$2^n(2^n-1)\equiv -2\equiv 5\bmod 7$.
Now notice that $2^n\bmod 7$ repeats with period $3$, so $2^n(2^n-1)$ also repeats with period three.
In particular it goes: $2,5,0,2,5,0\dots$
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