Solution:

The only solution pair is $(k ; n)=(2 ; 4)$. For every solution pair, $k^{4}+n^{2}$ and $7^{k}-3^{n}$ are integers, where $7^{k}-3^{n}$ need not necessarily be positive. Since $7^{k}$ and $3^{n}$ are odd, $2 \mid 7^{k}-3^{n}$ holds, and therefore $2 \mid k^{4}+n^{2}$. Hence $k$ and $n$ must be either both even or both odd. In the latter case $k^{4}+n^{2} \equiv 2 \bmod 4$ holds, whereas $7^{k}-3^{n} \equiv(-1)^{k}-(-1)^{n} \equiv 0 \bmod 4$. Therefore $k$ and $n$ are both even, and we set $k=2 a$ resp. $n=2 b$ with positive integers $a, b$. It follows that $7^{2 a}-3^{2 b} \mid(2 a)^{4}+(2 b)^{2}$ and further $49^{a}-9^{b} \mid 16 a^{4}+4 b^{2}$. Now $49^{a}-9^{b} \equiv 1^{a}-1^{b} \equiv 0 \bmod 8$, so $16 a^{4}+4 b^{2}$ must also be divisible by 8.

Therefore $b$ must be even and can be represented as $b=2 c$ with a suitable positive integer $c$. It follows that $7^{2 a}-9^{2 c}=(7^{a}+9^{c})(7^{a}-9^{c}) \mid 16(a^{4}+c^{2})$. Since always $7^{a} \neq 9^{c}$ and both powers are odd, $|7^{a}-9^{c}| \geq 2$, so that $7^{a}+9^{c} \mid 8(a^{4}+c^{2})$ must hold.

Lemma 1: For all $a \geq 4$, $7^{a}>8 a^{4}$.

Proof by complete induction on $a$:
For $a=4$, $2401=7^{4}>8 \cdot 4^{4}=2^{11}=2048$ is true. Now assume $7^{a}>8 a^{4}$.
Then $7^{a+1}=7 \cdot 7^{a}>7 \cdot 8 a^{4}=8(a+1)^{4} \cdot \frac{7 a^{4}}{(a+1)^{4}}>8(a+1)^{4} \cdot 7 \cdot\left(\frac{4}{5}\right)^{4}>8(a+1)^{4}$.

Lemma 2: For all $c \geq 1$, $9^{c}>8 c^{2}$.

Proof by complete induction on $c$:
For $c=1$, $9>8$ is true. Now assume $9^{c}>8 c^{2}$. Then
$9^{c+1}=9 \cdot 9^{c}>9 \cdot 8 c^{2}=8(c+1)^{2} \cdot \frac{9 c^{2}}{(c+1)^{2}}>8(c+1)^{2} \cdot 9 \cdot\left(\frac{1}{2}\right)^{2}>8(c+1)^{2}$.

From the lemmas it follows that for $a \geq 4$, because of $7^{a}+9^{c}>8(a^{4}+c^{2})$, no required numbers $k$ and $n$ exist. So only $a=1,2,3$ still need to be examined.

Lemma 3: For all $c \geq 3$, $9^{c}>305+8 c^{2}$.

Proof by complete induction on $c$:
For $c=3$, $9^{3}=729>305+8 \cdot 3^{2}=377$ is true. Now assume $9^{c}>305+8 c^{2}$. Then $9^{c+1}>9\left(305+8 c^{2}\right)=305+8\left(305+9 c^{2}\right)$. Because of
$305+9 c^{2}=(c+1)^{2}+8 c^{2}-2 c+304>(c+1)^{2}+c(c-2)>(c+1)^{2}$ for $c>2$, the claim follows.

Case 1: $a=1$. For $c=1$, $49-81=-32 \mid 32=16(1+1)$ holds. Therefore $(k ; n)=(2 ; 4)$ is a solution. For $c=2$, $7+81=88 \nmid 8(1+4)=40$ holds. For $c>2$, $7+9^{c} \leq 8+8 c^{2}$ must hold, hence $9^{c}<1+8 c^{2}$. This contradicts Lemma 3.

Case 2: $a=2$. For $c=1$, $7^{2}+9^{1}=58$ and $8\left(2^{4}+1^{2}\right)=136$ hold. For $c=2$, $49+81=130 \nmid 8(16+4)=160$ holds. For $c>2$, $49+9^{c} \leq 8\left(16+c^{2}\right)$ must hold, hence $9^{c} \leq 79+8 c^{2}$. This contradicts Lemma 3.

Case 3: $a=3$. For $c=1$, $343+9=352$ and $656=8\left(3^{4}+1\right)$ hold. For $c=2$, $343+81=424$ and $680=8\left(3^{4}+4\right)$ hold. For $c>2$, $343+9^{c} \leq 8\left(81+c^{2}\right)$ must hold, hence $9^{c} \leq 305+8 c^{2}$. This contradicts Lemma 3.