Consider a $k \times k$ array, where $k = 3^n$ for a positive integer $n$. The array contains each of the integers $1, 2, \ldots, m$ exactly once, and the remaining entries are all zeros. We are tasked with finding the smallest possible value of $m$ such that all row sums and column sums are equal.

In a $k \times k$ array with equal row and column sums $S$, the total sum of the entries is $k \times S$. Since the entries $1, 2, \ldots, m$ appear exactly once, the total sum of non-zero entries is:

$$\sum_{i=1}^{m} i = \frac{m(m+1)}{2}$$

To satisfy that the row sums and column sums are equal, the non-zero entries must be distributed such that their sum for any row or column leads to an integer average. This implies:

$$k \times S = \frac{m(m+1)}{2}$$

Given that $k = 3^n$, we analyze how to distribute the integers optimally to achieve the same row and column sums. We focus on ensuring each sum is the same while minimizing $m$.

Observing that the simplest scenario would involve filling entries up to the largest non-zero integer across rows or columns, we deduce that filling in consecutive numbers maximizes the use of non-zero entries uniformly across rows and columns:

Set $m = 3^{n+1} - 1$. This setting ensures that all $m$ non-zero numbers produce a sum that aligns with the requisite uniformity for both rows and columns:

- The total number of non-zero cells is $m = 3^{n+1} - 1$.
- This arises as the maximum integer sum obtainable for completed non-zero fills which ensures all sums equate.

Through careful arrangement, we achieve consistent row and column summations with the structure:
$$m = 3^{n+1} - 1$$

Thus, the smallest possible value of $m$ is:
$$\boxed{3^{n+1} - 1}$$