Factor a bivariate polynomial treating one variable as constant

Although there might be various approaches, it seems that proceeding the most obvious one should be good enough

We have

Collect[(x^2 - f[k]) (x^2 - g[k]) // Expand, x]

x^4 + x^2 (-f[k] - g[k]) + f[k] g[k]

on the other hand it should be identically equal to

Collect[4096 - 20480 k^2 + 16384 k^4 - 128 x^2 - 320 k^2 x^2 + x^4, x]

4096 - 20480 k^2 + 16384 k^4 + (-128 - 320 k^2) x^2 + x^4

Therefore equating appropriate coefficients in the both polynomials we get a simple system and solving it yields solutions

Solve[{-128 - 320 k^2 == -f[k] - g[k], 
       4096 - 20480 k^2 + 16384 k^4 == f[k] g[k]}, {f[k], g[k]}]

 {{f[k] -> 32 (2 + 5 k^2 - Sqrt[40 k^2 + 9 k^4]), 
   g[k] -> 32 (2 + 5 k^2 + Sqrt[k^2 (40 + 9 k^2)])}, 
  {f[k] -> 32 (2 + 5 k^2 + Sqrt[40 k^2 + 9 k^4]), 
   g[k] -> 32 (2 + 5 k^2 - Sqrt[k^2 (40 + 9 k^2)])}}

Another, and a slightly simpler approach would be:

Solve[ 
  PolynomialQuotientRemainder[
    4096 - 20480 k^2 + 16384 k^4 - 128 x^2 - 320 k^2 x^2 + x^4, 
     x^2 - f[k], x] == {x^2 - g[k], 0}, 
  {f[k], g[k]}]

Surely there's a simple, one-line solution for my problem?

Of course.

SolveAlways[16384 k^4 - 20480 k^2 + x^4 - 320 k^2 x^2 - 128 x^2 + 4096 ==
            (x^2 - f) (x^2 - g), x]
   {{f -> 32 (2 + 5 k^2 - Sqrt[k^2 (40 + 9 k^2)]), 
     g -> 32 (2 + 5 k^2 + Sqrt[40 k^2 + 9 k^4])},
    {f -> 32 (2 + 5 k^2 + Sqrt[k^2 (40 + 9 k^2)]), 
     g -> 32 (2 + 5 k^2 - Sqrt[40 k^2 + 9 k^4])}}