Let $d(n)$ be the number of divisors of $n$. Prove that we can colour the natural numbers using $2$ colours such that if for an infinite increasing sequence $(a_1,a_2,a_3,\cdots)$ the sequence $(d(a_1),d(a_2), \cdots)$ is non constant geometric progression, then all the terms $(a_1,a_2,a_3,\cdots)$ cannot have same colour. (You may use the fact that we can colour the natural numbers using $2$ Colours such that all the terms of any infinite increasing A. P. (Arithmatic Progression) cannot have same colour .)
I felt like we have to prove that the infinite sequence told in the question must be in AP but I got no way of proving that if the number of divisors are in GP then the terms should be in AP.
Take a function that grows very rapidly in $n$ ... $2^{n^2}$ for instance.
Now, color all the numbers with $d(n)=2=2^{1^2}$ red. Next color all the numbers with $2^{1^2}<d(n)≤ 2^{2^2}=16$ blue. Continue in this way, so those $n$ with $2^{2^2}<d(n)≤2^{3^2}$ are red and so on. The "stripes" are getting wider and wider.
Suppose your geometric series has ratio $r>1$. Choose $n$ large enough so that $2^{n}>r$ and consider the largest index $i$ such that $d(a_i)≤2^{n^2}$. We note that $d(a_i)>2^{(n-1)^2}$ since otherwise we'd have $$d(a_{i+1})=rd(a_i)<2^n\times 2^{(n-1)^2}<2^{n^2}$$ Then the colors of $a_i,a_{i+1}$ can not be the same since $$2^{n^2}<d(a_{i+1})=rd(a_i)<2^n\times 2^{n^2}=2^{n^2+n}<2^{(n+1)^2}$$
Note: your idea regarding progressions is not valid. For any prime $p$ we have $d(p^k)=k+1$ so choose a sequence $\{k_i\}$ such that the $(k_i+1)'s$ are in geometric progression and define $a_i=2^{k_i}$. It is clear that these $a_i$ are not in arithmetic progression.