I claim $n=1,5$ are the only answers. Verify it for $n\le 5$, suppose $n\ge 6$. Note that $6^n+1\equiv 1\pmod{2}$ and $\equiv 2\pmod{5}$, yielding its last digit to be $7$. Hence, $6^n+1=\overline{77\cdots 7}$. Letting $N$ to be the number of 7's, we get $N\ge 5$ and arrive at \[ 6^n +1 = 7\cdot \frac{10^N-1}{9}\iff 9\cdot 6^n + 16 = 7\cdot 10^N \iff 7\cdot 10^N = 16\left(9\cdot 3^n\cdot 2^{n-4}+1\right)\triangleq T. \]Clearly $v_2(T)=4$ whereas $N\ge 5$ as claimed above. This is a contradiction.