Since $m$ is a perfect number, it must be of the form $m=2^{k-1}(2^k-1)$ for some positive integer $k$. Therefore, we have

$$m = 2^{k-1}(2^k-1) = 3^n p.$$

Since $p$ is prime and $p\neq 2$, we have two cases:

Case 1: $p\equiv 1\pmod 3$

In this case, we have $3^n \mid 2^{k-1}$, which implies $k-1\geq n$. Furthermore, since $p\equiv 1\pmod 3$, we have $p+1\equiv 2\pmod 3$, which implies $3\mid p+1$. Thus, we have

$$m = 2^{k-1}(2^k-1) \geq 2^{n}(2^{n+1}-1) > 2^n(2\cdot 3^n -1) = 2p,$$

which contradicts the fact that $m$ is perfect. Therefore, this case cannot occur.

Case 2: $p\equiv 2\pmod 3$

In this case, we have $3^n \mid 2^k-1$, which implies $k\geq n+1$. Furthermore, since $p\equiv 2\pmod 3$, we have $p+1\equiv 0\pmod 3$, which implies $3^{n+1}\mid p+1$. Thus, we have

$$m = 2^{k-1}(2^k-1) \geq 2^{n}(2^{n+1}-1) = 3^n(3^{n+1}-2) > 3^n(3^n\cdot 2 – 2) = 2\cdot 3^{2n} > 2p,$$

which again contradicts the fact that $m$ is perfect. Therefore, this case cannot occur either.

Therefore, there is no prime $p\neq 2$ that can make $m$ a perfect number.

Note: If $p=2$, then $m=2^{2n+1}-2^n$, and it is known that this is a perfect number if and only if $2^n$ is an odd prime (known as Mersenne prime).

Since $m$ is perfect, we know that the sum of its divisors (excluding itself) is equal to $m$. That is, if $d_1,d_2,\ldots,d_k$ are the positive divisors of $m$ (excluding $m$ itself), then we have $$d_1+d_2+\cdots+d_k=m.$$ Now let $d$ be any positive divisor of $m$. Then $d$ can be written as $3^ap^b$ for some nonnegative integers $a$ and $b$ with $0\leq b\leq 1$. (Note that $b=0$ is possible because $m$ need not be odd.) Since $d$ divides $m$, we have $3^ap^b\mid 3^n p$, which implies $3^a\mid 3^n$. Therefore, $a\leq n$. It follows that $d\leq 3^n p$.

Now consider the sum of the divisors of $m$ that are divisible by $3$. These are precisely the divisors of the form $3^ap$ with $1\leq a\leq n$. There are $n$ such divisors, each of which is at most $3^n p$, so the sum of these divisors is at most $n\cdot 3^n p$. Therefore, the sum of the divisors of $m$ that are not divisible by $3$ is at least $$m-(n\cdot 3^n p)=(3^n- n\cdot 3^n)p.$$ Since $m$ is perfect, this sum must be equal to $m$. That is, $$(3^n-n\cdot 3^n)p=3^n p+2^k,$$ where $k$ is a nonnegative integer and $2^k$ is the sum of the divisors of $m$ that are equal to $2^j$ for some $j\geq 1$. Since $p$ is prime, it follows that $p$ divides $2^k$. However, $p$ is odd because $p\neq 2$, so $p$ and $2$ are relatively prime. Therefore, we must have $p=2$.

Let $d(m)$ denote the sum of the divisors of $m$. If $m$ is perfect, then $d(m)=2m$.

Since $m=3^n p$, the sum of the divisors of $m$ is given by $(1+3+\cdots+3^n)(1+p)$, which is equal to $\frac{3^{n+1}-1}{2}(p+1)$.

Thus, we have the equation $\frac{3^{n+1}-1}{2}(p+1)=2(3^n p)$.

Rearranging terms, we get $3^{n+1}p-3^n-1=0$.

Now, we observe that if $n\geq 2$, then $3^{n+1}p-3^n-1 \equiv 2 \pmod{3}$, which means that $3^n p$ cannot be perfect. Therefore, we must have $n=1$.

Plugging in $n=1$, we get $3p-2=0$, which implies $p=2/3$, a contradiction since $p$ is prime.

Therefore, there is no such $m$ such that $m=3^n p$ and $m$ is perfect.