6. If $2n+1, 20n+1 \left(n \in \mathbf{N}_{+}\right)$ are powers of the same positive integer, then all possible values of $n$ are
6. 4 .

According to the problem, we know that $(2 n+1) \mid(20 n+1)$. Then $(2 n+1) \mid[10(2 n+1)-(20 n+1)]=9$. Therefore, $n \in\{1,4\}$. Upon verification, $n=4$.