6. The number of positive integers $n$ that make $3^{n}+81$ a perfect square is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
6.B.

When $n \leqslant 4$, it is easy to see that $3^{n}+81$ is not a perfect square. Therefore, let $n=k+4\left(k \in \mathbf{N}_{+}\right)$. Then
$$
3^{n}+81=81\left(3^{k}+1\right) \text {. }
$$

Since $3^{n}+81$ is a perfect square and 81 is a square number, there must exist a positive integer $x$ such that
$$
3^{k}+1=x^{2},
$$

which means $3^{k}=x^{2}-1=(x+1)(x-1)$.
Thus, $x+1$ and $x-1$ are both powers of 3.
Since the two numbers $x+1$ and $x-1$ differ by 2, they can only be 3 and 1. Therefore, $x=2, k=1$.

Hence, there exists a unique positive integer $n=k+4=5$ such that $3^{n}+81$ is a perfect square.