An idea may be to find $A$ and $B$ prime such as : $A \mid 10^n+1$ $A^2 \nmid 10^n+1$ $B^2 \mid 10^n+1$ $A^3 < B^2$ Then let $N = A^2 * (10^n+1) / B^2$ then $N * (10^n+1)$ is a double number. After some tries, it happens that $A=7$ and $B = 19$ works and that (using modulo, Euler's theorem, ... to prove it) with $n = 171*(1+14*k)$

Using Euler's theorem and/or brute force:- $10^{21} \equiv -1 (mod 49) \\$ and $10^{42} \equiv 1 (mod 49) \\$ so I think that if $A = 16.(10^{42k+21} + 1)/49$ , then $A.(10^{42k+21} + 1)$ is square, and $A$ is also large enough to prevent leading zeroes (i.e. $A > 10^{42k+20}$ )

Observe that: $ 1001=7\cdot143\Rightarrow 10^3=-1+7a$, $ (a=143).$ Using binomial theorem we get that : $ 10^{21}=(-1+7a)^7=-1+7b^2$ for some integer $ b$, so that we also have $ 10^{21n}\equiv-1(\mod 49)$ for any odd integer $ n>0.$ Hence $ N=\frac{9}{49}(10^{21n}+1)$ is an integer of $ 21n$ digits, and ${ N(10^{21n}+1)=(\frac{3}{7}\cdot(10^{21n}+1}))^2$ is a double number that is a perfect square!