Infinite thermal current noise in a wire?
This seems strange! I understand that the final noise power does not depend on resistance, but still infinite noise density seems absurd.


No, it's neither strange nor absurd because you are dividing 0 by 0:

You get power from current by squaring and multiplying by R, so you get one R in the numerator and one in the denominator and both cancel out:

\$\require{cancel}P = \Delta f(nd_I)^2R = \Delta f \sqrt{\frac{k_B T}{R}}^2 R = \Delta f\frac{k_B T}{\cancel{R}} \cancel{R} = \Delta fk_BT\$
wich is independent of R.

So even if current noise density is infinite, noise power is not.