62nd Ukrainian National Mathematical Olympiad

Obviously, $x>y$ (since $x+3^y\ne y+3^x$). 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: $$1=y+3^x-x-3^y=3^y(3^{x-y}-1)-(x-y)\ge 1\cdot ((x-y)+2-1)-(x-y)=1.$$ 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 $$3^x-3^y=x-y-1\Rightarrow 3^y(3^{x-y}-1)=x-y-1\Rightarrow 3^{x-y}-1\le x-y-1\Rightarrow 3^{x-y}\le x-y$$ 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.