Suppose that the positive integer $n$ is such that

$$p^{n}-(p-1)^{n}=3^{a}$$

for some prime $p$ and positive integer $a$.

If $p=2$, then $2^{n}-1=3^{a}$ by $(1)$, whence $(-1)^{n}-1 \equiv 0(\bmod 3)$, so $n$ should be even. Setting $n=2 s$ we obtain $\left(2^{s}-1\right)\left(2^{s}+1\right)=3^{a}$. It follows that $2^{s}-1$ and $2^{s}+1$ are both powers of 3, but since they are both odd, they are co-prime, and we have $2^{s}-1=1$, i.e. $s=1$ and $n=2$. If $p=3$, then (1) gives $3 \mid 2^{n}$, which is impossible.

Let $p \geq 5$. Then it follows from (1) that we can not have $3 \mid p-1$. This means that $2^{n}-1 \equiv 0$ $(\bmod 3)$, so $n$ should be even, and let $n=2 k$. Then

$$p^{2 k}-(p-1)^{2 k}=3^{a} \Longleftrightarrow\left(p^{k}-(p-1)^{k}\right)\left(p^{k}+(p-1)^{k}\right)=3^{a}$$

If $d=\left(p^{k}-(p-1)^{k}, p^{k}+(p-1)^{k}\right)$, then $d \mid 2 p^{k}$. However, both numbers are powers of 3, so $d=1$ and $p^{k}-(p-1)^{k}=1, p^{k}+(p-1)^{k}=3^{a}$.

If $k=1$, then $n=2$ and we can take $p=5$. For $k \geq 2$ we have $1=p^{k}-(p-1)^{k} \geq p^{2}-(p-1)^{2}$ (this inequality is equivalent to $p^{2}\left(p^{k-2}-1\right) \geq(p-1)^{2}\left((p-1)^{k-2}-1\right)$, which is obviously true). Then $1 \geq p^{2}-(p-1)^{2}=2 p-1 \geq 9$, which is absurd.

It follows that the only solution is $n=2$.

To solve the problem, we need to find all positive integers $n$ such that there exists a prime number $p$ for which $p^n - (p-1)^n$ is a power of 3. Let's go through the solution step-by-step.

1. Initial Considerations:
- We are given that $p^n - (p-1)^n$ is a power of 3, i.e., $p^n - (p-1)^n = 3^a$ for some positive integer $a$.
- Since $3^0 = 1$ is not considered a power of 3, we have $n > 1$.

2. Modulo 3 Analysis:
- Consider the expression modulo 3. If $p \equiv 0 \pmod{3}$, then $p$ is divisible by 3, which is not possible since $p$ is a prime number.
- If $p \equiv 1 \pmod{3}$, then $p-1 \equiv 0 \pmod{3}$, making $(p-1)^n \equiv 0 \pmod{3}$. Thus, $p^n \equiv 3^a \pmod{3}$, which is not possible since $p^n \equiv 1 \pmod{3}$ and $3^a \equiv 0 \pmod{3}$.
- Therefore, $p \equiv 2 \pmod{3}$.

3. Simplifying the Expression:
- Given $p \equiv 2 \pmod{3}$, we have $p-1 \equiv 1 \pmod{3}$.
- Thus, $p^n \equiv 2^n \pmod{3}$ and $(p-1)^n \equiv 1^n \equiv 1 \pmod{3}$.
- Therefore, $p^n - (p-1)^n \equiv 2^n - 1 \equiv 0 \pmod{3}$, which implies $2^n \equiv 1 \pmod{3}$.

4. **Determining $n$:**
- The congruence $2^n \equiv 1 \pmod{3}$ holds if and only if $n$ is even. Let $n = 2k$.

5. **Further Analysis for $n = 2k$:**
- Consider $n = 4m$ (i.e., $n$ is a multiple of 4). We have:
$$p^4 - (p-1)^4 = (p^2 - (p-1)^2)(p^2 + (p-1)^2)$$
- Simplifying, we get:
$$p^2 - (p-1)^2 = p^2 - (p^2 - 2p + 1) = 2p - 1$$
$$p^2 + (p-1)^2 = p^2 + (p^2 - 2p + 1) = 2p^2 - 2p + 1$$
- Since $p \equiv 2 \pmod{3}$, we have:
$$p^2 \equiv 4 \equiv 1 \pmod{3} \quad \text{and} \quad (p-1)^2 \equiv 1 \pmod{3}$$
- Thus, $p^2 + (p-1)^2 \equiv 1 + 1 \equiv 2 \pmod{3}$, which cannot be a power of 3.

6. **Conclusion for $n = 2k$:**
- Since $n = 4m$ is not possible, we must have $n = 2k$ where $k$ is odd.
- If $k > 1$, then $p^k - (p-1)^k$ must also be a power of 3, which is not possible by our previous analysis.
- Therefore, $k = 1$, and $n = 2$.

7. Verification:
- For $n = 2$, we have:
$$p^2 - (p-1)^2 = p^2 - (p^2 - 2p + 1) = 2p - 1$$
- We need $2p - 1 = 3^a$. For $a = 1$, we get $2p - 1 = 3 \implies 2p = 4 \implies p = 2$, which is a prime number.
- For $p = 2$, we have $2^2 - 1^2 = 4 - 1 = 3 = 3^1$, which is indeed a power of 3.

Thus, the only solution is $n = 2$.

The final answer is $\boxed{n = 2}$.