Unfortunately, for sufficiently large numbers $x$, the remainder of $x$ modulo $m$ has absolutely nothing to do with the remainder of $x$ modulo $n$ when $m$ and $n$ are relatively prime.
It may be possible to use some method to use the remainder modulo $2^{20}$ to accelerate the calculation of the remainder modulo $2^{96}$ (which would be the exact value) and then reduce that result modulo $2^{20} + 1$. For this particular calculation, I find it very unlikely that any such approach would be more efficient than computing it directly.