Solution:

Let $y = \sqrt[16]{5}.$ Multiplying the numerator and denominator by $y - 1$ telescopes the denominator by repeated difference of squares: $$x = \frac{4(y-1)}{(y^8+1)(y^4+1)(y^2+1)(y+1)(y-1)} = \frac{4(y-1)}{y^{16} - 1} = \frac{4(y-1)}{4} = y - 1.$$

Hence $x + 1 = y = 5^{1/16},$ and $(x+1)^{48} = 5^{48/16} = 5^3 = 125.$