Answer: $63$

Use $n^2-1=(n+1)(n-1)$ for all values of $n$

$2^{48}=\left(2^{24}\right)^2$, so $$\begin{align}

2^{48}-1&=\left(2^{24}\right)^2-1\\

&=\left(2^{24}+1\right)\left(2^{24}-1\right)\\

&=\left(2^{24}+1\right)\left(\left(2^{12}\right)^2-1\right)\\

&=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(2^{12}-1\right)\\

&=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(\left(2^{6}\right)^2-1\right)\\

&=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(2^{6}+1\right)\left(2^{6}-1\right)\\

\end{align}$$

$2^4=16, 2^5=32, 2^6=64$

$2^{48}-1=\left(2^{24}+1\right)\left(2^{12}+1\right)\ \underbrace{\left(2^{6}+1\right)}_{65}\ \underbrace{\left(2^{6}-1\right)}_{63}$

The smaller one is $63$