Step 1: Recall a useful identity.
For positive numbers $a$ and $b$, there is a neat identity: $a^{\log_{10}(b)} = b^{\log_{10}(a)}$. This comes from writing $a = 10^{\log_{10}(a)}$ and $b = 10^{\log_{10}(b)}$ and comparing the exponents once both sides are expressed with base $10$.
Step 2: Apply the identity directly.
Here $x = n^{\log_{10}(m)}$ is exactly of the form $a^{\log_{10}(b)}$ with $a = n$, $b = m$. By the identity above, this equals $m^{\log_{10}(n)}$.
But $m^{\log_{10}(n)}$ is exactly how $y$ is defined.
Step 3: Conclude the relation.
So $x = n^{\log_{10}(m)} = m^{\log_{10}(n)} = y$, giving $x = y$ directly, without even needing to plug in numbers.
Step 4: Sanity-check with a number example.
Let $m = 1000$ and $n = 10$ (both positive, $m > n$).
$x = 10^{\log_{10}(1000)} = 10^{3} = 1000$.
$y = 1000^{\log_{10}(10)} = 1000^{1} = 1000$.
Both give $1000$, confirming $x = y$ for this pair too, and the relation does not depend on which valid $m, n$ we pick.
Step 5: Rule out the wrong options.
Since $x$ and $y$ always come out equal, neither "$x > y$" nor "$x < y$" can be a general rule; they would only look true or false by accident for one particular chosen pair. Option (D), $x = \log_{10}(y)$, confuses the value of $x$ with the logarithm of $y$; since $x = y$ here, that option would need $x$ to equal $\log_{10}(x)$, which fails for a value like $1000$.
Step 6: Final Answer.
\[ \boxed{x = y} \]