I have been given the sequence of numbers
$$a(n) = 2^{2n-1}-n$$
and I want to know if any of its members is a power of 10 (other than 1). The only thing I've figured out is that the $n$ for such a number must be even and not a multiple of five. By searching, I know that there are no powers of 10 in the first 10000 terms, but I don't know how to prove that there are none, if that is the case.
Consider that, for some integer $n \ge 2$, there's a positive integer $k$ where
$$2^{2n-1} - n = 10^k = 2^k\times 5^k \tag{1}\label{eq1A}$$
Since $2n - 1 \gt k$ so $2^k \mid 2^{2n-1}$, then with $2^k \mid 10^k$, we also have $n = 2^{k}m$ for some integer $m$. Dividing both sides of \eqref{eq1A} by $2^{k}$ gives
$$2^{2^{k+1}m - 1 - k} - m = 5^k \tag{2}\label{eq2A}$$
No $n \le 12$ works, so consider only $n \gt 12$. With $2^{23} - 12 = 8\,388\,596$, this requires $k \ge 7$. We therefore then also get
$$\begin{equation}\begin{aligned} 2^{2^{k+1}m - 1 - k} - m & \ge 2^{2^{k+1} - 1 - k} - 1 \\ & \gt 2^{2^k} \\ & = 16^{2^{k-2}} \\ & \gt 16^{k} \\ & \gt 5^{k} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
However, this contradicts \eqref{eq2A}, showing there are no solutions to \eqref{eq1A} for $n \ge 2$.