Highest power of $2$

Note that $5^{2^n} - 1 \; = \; \big(5^{2^{n-1}}-1\big)(5^{2^{n-1}} +1\big) \hspace{2cm} n \in \mathbb{N}$ It is clear that $5^m \equiv 1 \pmod{4}$ , so that $5^m + 1 \equiv 2 \pmod{4}$ for all $m \in\mathbb{N}$ . Thus, if $I_2(k)$ is the index of $2$ in $k$ , then $I_2\big(5^{2^n}-1\big) \; = \; I_2\big(2^{2^{n-1}}-1\big) + 1 \hspace{2cm} n \in \mathbb{N}$ and hence it follows that $I_2\big(5^{2^n}-1\big) \; = \; I_2(5^2-1) + n-1 \; = \; n+2 \hspace{2cm} n \in \mathbb{N}$ With $n=2019$ , we obtain the answer $\boxed{2021}$ .