Let $k = \log_2(3^n + 7)$, where $k$ is a positive integer. Then $3^n + 7 = 2^k$.

So $2^k - 3^n = 7$.

We seek all positive integers $n$ such that this equation has a solution in positive integers $k$.

Let us try small values of $n$:

For $n = 1$: $2^k - 3 = 7 \implies 2^k = 10$ (not a power of $2$).

For $n = 2$: $2^k - 9 = 7 \implies 2^k = 16 \implies k = 4$.
So $n = 2$ works.

For $n = 3$: $2^k - 27 = 7 \implies 2^k = 34$ (not a power of $2$).

For $n = 4$: $2^k - 81 = 7 \implies 2^k = 88$ (not a power of $2$).

For $n = 5$: $2^k - 243 = 7 \implies 2^k = 250$ (not a power of $2$).

For $n = 6$: $2^k - 729 = 7 \implies 2^k = 736$ (not a power of $2$).

Let us check if there are any other solutions for larger $n$.

Suppose $n \geq 2$. Then $3^n$ grows much faster than $2^k$ for large $n$ unless $k$ is also large. But $2^k = 3^n + 7 > 3^n$, so $2^k > 3^n$.

But $2^k = 3^n + 7 < 2 \cdot 3^n$ for large $n$ (since $7$ is negligible compared to $3^n$), so $2^k < 2 \cdot 3^n$.

Thus, $3^n < 2^k < 2 \cdot 3^n$.

Take logarithms:

$n \log 3 < k \log 2 < n \log 3 + \log 2$

So $k$ is approximately $n \log_2 3$.

But $\log_2 3 \approx 1.58496$, so $k \approx 1.58496 n$.

But $2^k - 3^n = 7$ is a very restrictive equation. For large $n$, $2^k$ and $3^n$ are both huge, and their difference is $7$.

Let us check for $n = 0$ (but $n$ must be positive), so skip.

Let us check for $n = 2$ (already found), $k = 4$.

Suppose $k = 3$, $2^3 = 8$, $3^n = 1$ (no integer $n$).

Suppose $k = 5$, $2^5 = 32$, $3^n = 25$ (not a power of $3$).

Suppose $k = 6$, $2^6 = 64$, $3^n = 57$ (not a power of $3$).

Suppose $k = 7$, $2^7 = 128$, $3^n = 121$ (not a power of $3$).

Suppose $k = 8$, $2^8 = 256$, $3^n = 249$ (not a power of $3$).

Suppose $k = 9$, $2^9 = 512$, $3^n = 505$ (not a power of $3$).

Suppose $k = 10$, $2^{10} = 1024$, $3^n = 1017$ (not a power of $3$).

So, only $n = 2$ works.

Therefore, the only positive integer $n$ such that $\log_2(3^n + 7)$ is a positive integer is $n = 2$.