Obviously, $x > y$ (since $x + 3^y \neq y + 3^x$). First, we prove the following lemma.

Lemma. For any natural number $n > 1$, the inequality $3^n > n + 2$ holds.

*Proof.* We prove the statement by induction. For $n=2$, we have that $3^2 = 9 > 2 + 2 = 4$. Suppose that the statement holds for some $n \ge 2$. Then
$$3^{n+1} = 3 \cdot 3^n > 3 \cdot (n+2) = 3n + 6 > n + 3 = (n+1) + 2.$$
*Lemma proved.*

Note that the equality $3^n = n + 2$ is only achieved when $n = 1$.

Consider the difference:
$$1 = y + 3^x - x - 3^y = 3^y(3^{x-y} - 1) - (x-y) \geq 1 \cdot ((x-y) + 2 - 1) - (x-y) = 1.$$
For this equality to hold, all intermediate inequalities must also be equalities, so the following conditions must be satisfied: $3^y = 1$ and $x - y = 1$, which gives us the answer stated above.

If $1 = (x + 3^y) - (y + 3^x)$, then
$$3^x - 3^y = x - y - 1 \Rightarrow 3^y(3^{x-y} - 1) = x - y - 1 \Rightarrow 3^{x-y} - 1 \leq x - y - 1 \Rightarrow 3^{x-y} \leq x - y$$
But we have proved that for all natural numbers $n$, $3^n \ge n + 2 > n \Rightarrow 3^n > n$, so we obtain a contradiction in this case.