6. Solve the equation $2^{x}+63^{y}=z^{2}$ in natural numbers.
Answer: $(8,2,65)$.

Solution. Since the left side of the equality is odd and not divisible by 3, the number $z$ is also odd and not divisible by 3. Then $z=6u \pm 1$ and $z^{2}=36u^{2} \pm 12u + 1$, which means $z^{2}$ gives a remainder of 1 when divided by 3 and a remainder of 1 when divided by 4. Note that the remainders of powers of 2 when divided by 3 alternate: for odd powers, it is 2, and for even powers, it is 1. Therefore, the number $x$ is even and, in particular, not less than two. Then $2^{x}$ is divisible by 4. Note that
the remainders of powers of 63 when divided by 4 also alternate: for odd powers, it is 3, and for even powers, it is 1. Since $z^{2}$ gives a remainder of 1 when divided by 4, the number $y$ is also even. Let $x=2k$ and $y=2n$. Then

$$
2^{2k}=z^{2}-63^{2n}=\left(z-63^{n}\right)\left(z+63^{n}\right)
$$

The product of the numbers $z-63^{n}$ and $z+63^{n}$ is a power of two, so each of them is also a power of two. Therefore, $z-63^{n}=2^{a}$ and $z+63^{n}=2^{b}$ for some non-negative integers $a<b$. Then $2^{b}-2^{a}=2 \cdot 63^{n}$ and therefore $2^{b-1}-2^{a-1}=63^{n}$. The right side of the equation is odd, so $a=1$ and, thus, $b-1=2k$. Therefore, $2^{2k}-1=63^{n}$. Thus,

$$
63^{n}=\left(2^{k}-1\right)\left(2^{k}+1\right)
$$

Since the numbers $2^{k}-1$ and $2^{k}+1$ are coprime (they differ by 2 and are odd) and their product is a power of 63, either $2^{k}-1=1$ and $2^{k}+1=63^{n}$, or $2^{k}-1=7^{n}$ and $2^{k}+1=9^{n}$. The first case is impossible, and in the second case, we get that $7^{n}+2=9^{n}$ and, thus, $n=1$. Therefore, $y=2n=2$ and $x=2k=8$. Substituting the found $x$ and $y$ into the left side, we get that $z=65$.