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 $n=3k+2$

The binomial coefficient is $2^n (2^n -1)/2 = 2^{n-1} (2\cdot 2^{n-1} -1)$. 
For the divisibility to hold you need this to be $-1 \pmod{7}$. 
Now the sequence nonnegative powers of $2$ modulo $7$ have a period of at most $6$, by Fermat's little theorem, but actually it is just has period $3$, and it starts
$2^0 = 1, 2^1 = 2, 2^2 =4, 2^3 =8\equiv 1, 2, 4, \dots$. 
So $2^{3k}\equiv 1$, $2^{3k+1}\equiv 2$, and $2^{3k+2}\equiv 4$  modulo $3$.
It remains to check if you have the desire congruence 
$$2^{n-1} (2\cdot 2^{n-1} -1) \equiv -1 \pmod{7}$$
somewhere. 
Since 
$$2 (2\cdot 2 -1) \equiv -1 \pmod{7}$$
Indeed you have it for $2^{n-1} \equiv 2 \pmod{7}$, and only for those, 
that is $n-1 \equiv 1 \pmod{3}$, which is just what you claim.   
