IMO!

Since $A = 7^k - 3^n$ is the difference of two odd numbers, it must be even. This implies that $B = k^4 + n^2$ must also be even, showing that $k$ and $n$ are of the same parity (both odd or both even).

First, see if $k$ and $n$ could be odd: say, $k = 2h+1$ and $n = 2m+1$ . We have $A = 7\cdot (48+1)^h - 3\cdot (8+1)^m = (8c_1+7) - (8c_2+3) = 4(2c_3+1).$ $B = (8c + 1) + (4m^2 + 4m + 1) = 2(4c + 2m(m+1) + 1) = 2(4c' + 1).$ But if $A|B$ then $4|2$ , which is impossible.

Therefore $k$ and $n$ must be even: $k = 2h$ , $n = 2m$ . Then $A = (48+1)^h - (8+1)^m = 8c,$ $B = 16h^4 + 4m^2 = 4(4h^4+m^2),$ $A|B\ \ \ \Longleftrightarrow\ \ \ 8c|4(4h^4 + m^2).$ This implies that $m$ is even, say $m = 2\ell$ so that $n = 4\ell$ . Now we have $A = 49^h - 81^\ell = (48+1)^h-(80+1)^\ell = 16c',$ $B = 16(h^4 + m^2).$ $A|B\ \ \ \Longleftrightarrow\ \ \ c'|h^4 + m^2.$

The smallest $h, m$ for which this can be true is $h = m = 1$ . This gives $k = 2, n= 4$ . Check: $7^2 - 3^4 = 49 - 81 = -32 | 32 = 16 + 16 = 2^4 + 4^2.$ I don't know if there are more solutions; for small $k, n$ certainly not. The divisor $7^k-3^n$ will be somewhat smallish if $n/k \approx \log 7/\log 3 \approx 1.7712$ , but whether it will be small enough to divide $k^4 + n^2$ is not certain.