1. 169 Determine all positive integer triples $(a, b, c)$ such that $a^{2}=2^{b}+c^{4}$ and only one of $a$, $b$, $c$ is odd.
[Solution] From $a^{2}=2^{b}+c^{4}$, we know $2^{b}=\left(a-c^{2}\right)\left(a+c^{2}\right)$.
Since $2^{b}$ is even, $a-c^{2}$ and $a+c^{2}$ have the same parity, so $a-c^{2}$ and $a+c^{2}$ are even. Without loss of generality, let
$$
\begin{array}{l}
a-c^{2}=2^{k}, \\
a+c^{2}=2^{k+m}.
\end{array}
$$

Here, $k$ and $m$ are positive integers. From (1) and (2), we know that $a$ and $c$ have the same parity. If $a$ and $c$ are both odd, it contradicts the given conditions, so $a$ and $c$ must both be even.
Subtracting (1) from (2) gives $2 c^{2}=2^{m+k}-2^{k}=2^{k}\left(2^{m}-1\right)$,

which simplifies to $c^{2}=2^{k-1}\left(2^{m}-1\right)$.
Since the left side is a perfect square and $2^{m}-1$ is odd, $2^{m}-1$ must also be a perfect square. Let
$$
2^{m}-1=(2 t-1)^{2}, t \in \mathbb{N}.
$$

Then $2^{m-1}=2 t^{2}-2 t+1$.
This equation holds only when $m=1$, so
$$
c^{2}=2^{k-1}, c=2^{\frac{k-1}{2}}.
$$
Since $c$ is a natural number and even, $k$ must be an odd number greater than 2. Let $\frac{k-1}{2}=n$, where $n$ is a positive integer. From (1),
$$
a=2^{k}+c^{2}=2^{2 n+1}+2^{2 n}=3 \cdot 2^{2 n},
$$

so
$$
\begin{aligned}
2^{b} & =\left(3 \cdot 2^{2 n}\right)^{2}-2^{4 n} \\
& =9 \cdot 2^{4 n}-2^{4 n}=2^{4 n+3}.
\end{aligned}
$$

Thus, $b=4 n+3$, and therefore
$$
(a, b, c)=\left(3 \cdot 2^{2 n}, 4 n+3,2^{n}\right)
$$

where $n$ is a positive integer, and only $b$ is odd.