6. The positive integer solutions of the equation $-m^{4}+4 m^{2}+2^{n} m^{2}+2^{n}+5=0$ are ( ) groups.
(A) 1
(B) 2
(C) 4
(D) Infinite
6. A.

From the factorization of the left side of the equation, we get
$$
\left(m^{2}+1\right)\left(-m^{2}+2^{n}+5\right)=0 \text {. }
$$

Therefore, $-m^{2}+2^{n}+5=0$.
So, $2^{n}+4=(m+1)(m-1)$.
Since $m+1$ and $m-1$ have the same parity, they are both even, which means $m$ is odd.
Let $m=2k+1$. Then $2^{n-2}+1=k(k+1)$.
Thus, $2^{n-2}$ is odd, so $n=2, m=3$.
Hence, there is 1 set of positive integer solutions.