Step 1: Test the relation with actual numbers first.
Pick two distinct positive numbers with $m > n$, say $m = 100$ and $n = 10$. Then:
\[
x = n^{\log_{10} m} = 10^{\log_{10}(100)} = 10^{2} = 100
\]
\[
y = m^{\log_{10} n} = 100^{\log_{10}(10)} = 100^{1} = 100
\]
Here $x = y = 100$, so for this pair the two quantities are equal.
Step 2: Try a second pair to make sure it is not a coincidence.
Take $m = 1000$, $n = 100$:
\[
x = 100^{\log_{10}(1000)} = 100^{3} = 1{,}000{,}000
\]
\[
y = 1000^{\log_{10}(100)} = 1000^{2} = 1{,}000{,}000
\]
Again $x = y$. Two different examples both giving equality is a strong hint that $x = y$ holds in general, not just by chance.
Step 3: Prove it holds for every m, n using logs.
Take $\log_{10}$ of both quantities. Using $\log_{10}(a^k) = k\log_{10}(a)$:
\[
\log_{10} x = \log_{10}(m) \cdot \log_{10}(n), \qquad \log_{10} y = \log_{10}(n) \cdot \log_{10}(m)
\]
These are the same product written in the two possible orders, so $\log_{10} x = \log_{10} y$, which forces $x = y$ since $\log_{10}$ never gives the same output for two different positive inputs.
Step 4: Note that the condition m > n never entered the proof.
Nowhere in Step 3 did the size comparison between $m$ and $n$ get used, only that both are positive. This tells us the equality $x = y$ would hold even if we swapped which of $m, n$ is larger, which is a good sign that options (A) and (B), the strict-inequality choices, are traps built around the given condition $m > n$ rather than being genuine consequences of it.
\[
\boxed{x = y}
\]