1. Find all pairs of positive integers $(k, n)$ such that $\left(7^{k}-3^{n}\right) \mid\left(k^{4}+n^{2}\right)$
1. $(2,4)$.

Assume the positive integer pair $(k, n)$ satisfies the condition.
Since $7^{k}-3^{n}$ is even, $k^{4}+n^{2}$ is also even. Therefore, $k$ and $n$ have the same parity.
If $k$ and $n$ are both odd, then
$$
k^{4}+n^{2} \equiv 1+1=2(\bmod 4).
$$

However, $7^{k}-3^{n} \equiv 7-3 \equiv 0(\bmod 4)$, which is a contradiction.
Therefore, $k$ and $n$ are both even.
Let $k=2a, n=2b$. Then
$7^{k}-3^{n}=7^{2a}-3^{2b}=\frac{7^{a}-3^{b}}{2} \cdot 2\left(7^{a}+3^{b}\right)$.
Since $\frac{7^{a}-3^{b}}{2}$ and $2\left(7^{a}+3^{b}\right)$ are both integers, we have
$2\left(7^{a}+3^{b}\right) \mid\left(7^{k}-3^{n}\right)$.
Also, $\left(7^{k}-3^{n}\right) \mid\left(k^{4}+n^{2}\right)$, which means
$\left(7^{k}-3^{n}\right) \mid 2\left(8a^{4}+2b^{2}\right)$,
thus $7^{a}+3^{b} \leqslant 8a^{4}+2b^{2}$.
We prove by mathematical induction:
When $a \geqslant 4$, $8a^{4}+2b^{2}$.
The above inequality cannot hold, so $a \leqslant 3$.
(1) When $a=1$, $k=2$.
From $8+2b^{2} \geqslant 7+3^{b}$, we get $2b^{2}+1 \geqslant 3^{b}$. This only holds when $b \leqslant 2$.

If $b=1$, then $n=2$, $\frac{k^{4}+n^{2}}{7^{k}-3^{n}}=\frac{2^{4}+2^{2}}{7^{2}-3^{2}}=\frac{1}{2}$ is not an integer.
If $b=2$, then $n=4$,
$$
\frac{k^{4}+n^{2}}{7^{k}-3^{n}}=\frac{2^{4}+4^{2}}{7^{2}-3^{4}}=-1.
$$

Therefore, $(k, n)=(2,4)$ is a solution.
(2) When $a=2$, $k=4$. Then
$k^{4}+n^{2}=256+4b^{2}$
$\geqslant\left|7^{4}-3^{n}\right|=\left|49-3^{b}\right|\left(49+3^{b}\right)$.
Since the minimum value of $149-3^{b}$ is 22, and it is achieved when $b=3$, we have $128+2b^{2} \geqslant 11\left(49+3^{b}\right)$, which contradicts $3^{b}>2b^{2}$.
(3) When $a=3$, $k=6$. Then
$k^{4}+n^{2}=1296+4b^{2}$
$\geqslant\left|7^{b}-3^{n}\right|=\left|343-3^{b}\right|\left(343+3^{b}\right)$.
Similarly, since $\left|343-3^{b}\right| \geqslant 100$, and it is achieved when $b=5$, we have $324+b^{2} \geqslant 25\left(343+3^{b}\right)$, which is a contradiction.
In conclusion, the solution that satisfies the condition is $(2,4)$.