1. In the set of integers, solve the equation

$$
x^{2}+y^{4}+1=6^{z}
$$
Solution. It is obvious that $z \geq 0$. If $z \geq 2$, then

$$
x^{2}+y^{4}+1 \equiv 0(\bmod 4)
$$

or

$$
x^{2}+y^{4} \equiv 3(\bmod 4)
$$

This is not possible because the remainders of squares of integers when divided by 4 are 0 or 1. Therefore, $0 \leq z<2$.

If $z=0$, then $x=y=0$.

If $z=1$, then $x^{2}+y^{4}=5$, or

$$
(x, y)=\{(2,1),(-2,1),(2,-1),(-2,-1)\}
$$

Thus,

$$
(x, y, z)=\{(0,0,0),(2,1,1),(-2,1,1),(2,-1,1),(-2,-1,1)\}
$$

are the solutions to the given equation.