Solve the following equation in the set of positive integer triples:

$$
1+2^{x}+3^{y}=z^{3} .
$$
Since the left side of the equation is always even, $z$ must also be even, so $z^{3}$ is definitely divisible by 8. Thus, $8 \mid 1+2^{x}+3^{y}$ holds. $3^{y}$ divided by 8 can only leave a remainder of 1 or 3, so $2^{x}$ divided by 8 must leave a remainder of 6 or 4. However, $2^{x}$ is a power of 2, so only the latter possibility can hold, and in this case $x=2$. Therefore, the equation can be reduced to $5+3^{y}=z^{3}$.

$z^{3}$ divided by 9 can leave a remainder of 0, 1, or 8, so $3^{y}$ cannot be divisible by 9. Therefore, only $y=1$ is possible, from which $z^{3}=5+3=8$, or $z=2$.

Thus, the only solution to the equation is $x=2, y=1, z=2$.