Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$? This question arose on the code golf StackExchange:

Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$?

In other terms: is there an integer that looks like $\color{blue}{1000...001000...}$ in both binary and decimal?
I feel like there probably isn't, but I can't think of a simple counterargument.
(A computer search by one of the commenters suggests there is no such integer up to $10^{100000}$.)

For a prime $p$, let $\nu_p(n)$ denote the exponent of $p$ in the prime factorization of $n$. We will use the Lifting the Exponent lemma.
Firstly, note that
$$a=\nu_2(2^a+2^b)=\nu_2(10^c+10^d)=c,$$
so we have $2^a+2^b=10^a+10^d$. Let $b=a+x$ and $d=a+y$, and write
$$1+2^x=5^a(1+10^y).\tag{1}\label{eq1}$$
We observe that
$$2^b<2^a+2^b<2\cdot 10^d\implies a+x< 1+(a+y)\log_2(10)<1+\frac 72a+\frac 72y,$$
so $x<1+\frac 52a+\frac 72y$, and
$$2^a+2^{a+x}=10^a+10^{a+y}\implies 2^{a+x}>10^{a+y}\implies a+x>a+y\implies x>y.$$
We see that, from (*),
$$5^a|1+2^x|2^{4x}-1\implies a\leq \nu_5(2^{4x}-1)=1+\nu_5(x).$$
On the other hand, looking at powers of $2$, we see that, modding out by $2^y$,
$$1\equiv 5^a\bmod 2^y,$$
which implies that
$$y\leq \nu_2(5^a-1)=1+\nu_2(a).$$
We now have the inequalities
$$y\leq 1+\nu_2(a),\ \ a\leq 1+\nu_5(x),\ \ x<1+\frac 52a+\frac 72y.$$
Since the functions on the right sides of the first two inequalities are very small, this is basically enough to finish -- we just need to actually carry out the bounding. We note that
$$\nu_p(x)\leq \log_p(x)=1+\log_p\left(\frac xp\right)<1+\frac{x}{p}.$$
As a result,
$$y<2+\frac{a}{2},\ a<2+\frac{x}{5},\ x<1+\frac 52a+\frac 72y.$$
This gives
$$a<2+\frac{2+5a+7y}{10}<2+\frac{2+5a}{10}+\frac{7}{10}\left(2+\frac a2\right)=\frac{18}5+\frac{17a}{20},$$
which gives that $a<24$. In addition, we have $y<2+a/2=14$ and $x<110$, so this is just a finite case check (which it seems like one of the commenters has performed to a satisfactory bound).
