I am looking at the question, whether for some positive integers $h,k,l\in\mathbb{N}$ we have that
$$\left(3^{h+l}-2^{2(k+h)+l}\right)\,\big| \left( 2^{2k+l+1}-2^2 3^{l-1}-2^{2k+1}3^l\right).$$
I conjecture by trying some examples, that the divisibility is not given if all $h,k,l$ are positive. The term $2^{2k+l+1}-2^2 3^{l-1}-2^{2k+1}3^l$ is always even and $3^{h+l}-2^{2(k+h)+l}$ is always odd, such that we have divisibility if we can write for some integer $N$ the first term as $$2^{2k+l+1}-2^2 3^{l-1}-2^{2k+1}3^l=2N(3^{h+l}-2^{2(k+h)+l})$$ and pulling out a power $2^2$ from the left hand side, we get $$2^{2k+l+1}-2^2 3^{l-1}-2^{2k+1}3^l=2^2(2^{2k+l-1}-3^{l-1}-2^{2k-1}3^l).$$ The term $2^{2k+l-1}-3^{l-1}-2^{2k-1}3^l$ is odd. Here I get stuck, because there is this dependency on $h$ in $3^{h+l}-2^{2(k+h)+l}$, which I can not seem to get rid off.
I tried also to conclude something by writing the whole expression as integer valued polynomials by setting $y_i = 3^{i-1}$, $i\in\lbrace h,l\rbrace $ and $x_j=2^j$, $j\in\lbrace k,h,l\rbrace$. This gives $p(x_k,x_l,x_h,y_h,y_k)=2x_k^2x_l-2^2y_l-6x_k^2y_l$ and $q(x_k,x_l,x_h,y_h,y_k)=9y_hy_l-x_k^2x_h^2x_l$, which are coprime as polynomials. But this does not necessarily mean, that there is no value $(a,b,c,s,t)$, at which they satisfy $\dfrac{p(a,b,c,s,t)}{q(a,b,c,s,t)}\in\mathbb{Z}$. So, I get stuck here...
Does anyone know how to proceed?
