The question is a bit strange, because it would also be true if one removes all the $2$ from the statement. Namely:
$$\Phi'(m, n):=m!^n \prod_{k = 1} ^ n \binom{m+k+x}{m}$$
is also symmetric.

Let me explain the symmetry of this modified function $\Phi'$, and you can certainly adapt it to your $\Phi$. But the explanation is somewhat artificial.

Assume harmlessly that $x$ is a non-negative integer. Consider a grid of size $m \times n$. The rows and columns are indexed $0, \cdots, m - 1$ and $0, \cdots, n - 1$, respectively, and a cell is indexed $(i, j)$ if it is in row $i$ and column $j$.

Now one wants to fill in the grid with integers, one in a cell, with the following restriction: every integer is at most $m + n + x$, and the integer in the cell $(i, j)$ should be strictly larger than $i + j$.

Here is the claim:

## For every $k \in \{0, \cdots, n - 1\}$, the number of different ways to fill in the column $k$ is equal to $$m!\binom{m + n - k + x}{m}.$$

This explains everything.

To prove the claim, it is (again) algebraically obvious, but if you really want a combinatorial proof, then there is the following bijection:

$$\{(a_1, \cdots, a_m): a_i \in \{1, \cdots, M\}, a_i \neq a_j \} \longleftrightarrow \{(c_1, \cdots, c_m): c_i \in \{1, \cdots, M - i\}\},$$

which is described as follows: given such $(a_i)_i$, define $c_i$ to be the integer such that $a_i$ is the $c_i$-th largest number in the set $\{1, \cdots, M\} \backslash \{a_1, \cdots, a_{i - 1}\}$. The procedure is obviously reversible, hence gives a bijection.