Let $x<y<z$. Solve the following equation in the set of natural numbers:

$$
3^{x}+3^{y}+3^{z}=179415
$$
I. Solution. The prime factorization of 179415 is: $3^{4} \cdot 5 \cdot 443$.

In the prime factorization, the number 3 appears to the fourth power. We factor out $3^{x}$ from the left side of the equation (where $x<y<z$):

$$
3^{x}\left(1+3^{y-x}+3^{z-x}\right)=179415
$$

Since the factor in parentheses is not divisible by 3, $x=4$. Thus:

$$
\begin{aligned}
3^{4}\left(1+3^{y-4}+3^{z-4}\right) & =179415 \\
1+3^{y-4}+3^{z-4} & =2215 \\
3^{y-4}+3^{z-4} & =2214
\end{aligned}
$$

The prime factorization of 2214 is: $2 \cdot 3^{3} \cdot 41$.

In this prime factorization, the number 3 appears to the third power. Since both terms on the left side of the equation can be factored by $3^{y-4}$, it is certain that $y-4=3$, so $y=7$. Thus:

$$
\begin{aligned}
3^{7-4}\left(1+3^{z-4-7+4}\right) & =2214 \\
3^{3}\left(1+3^{z-7}\right) & =2214 \\
3^{z-7} & =81
\end{aligned}
$$

Thus $z-7=4$, so $z=11$.

Therefore, the solution to the equation is: $x=4, y=7, z=11$.

II. Solution. Rewrite the number on the right side of the equation in base 3: 100010010000.

Considering the place values in base 3: $3^{4}+3^{7}+3^{11}=179415$. Since $x<y<z$, we have $x=4, y=7$, $z=11$. This is the only solution to the equation, as any number can be uniquely represented in any base.