Can the powers of 2 in the denominators of a fraction sum be used to find a contradiction?

Let ${a}_{1},{a}_{2},{b}_{1},{b}_{2},{c}_{1},{c}_{2},{d}_{1},{d}_{2},{e}_{1},{e}_{2}$ be odd integers, with

$gcd({a}_{1},{a}_{2})=gcd({b}_{1},{b}_{2})=gcd({c}_{1},{c}_{2})=gcd({d}_{1},{d}_{2})=gcd({e}_{1},{e}_{2})=1,$

and assume $n\ge 3$ and $m\ge 1$ are integers such that

$\begin{array}{}\text{(}\star \text{)}& \frac{{a}_{1}}{{2}^{3(n-1)m}{a}_{2}}+\frac{{b}_{1}}{{2}^{3((n-1)m+1)}{b}_{2}}+\frac{{c}_{1}}{{2}^{3((n-1)m+1)}{c}_{2}}+\frac{{d}_{1}}{{2}^{3nm}{d}_{2}}+\frac{{e}_{1}}{{2}^{3nm}{e}_{2}}=1.\end{array}$

Is there a way to prove, using only the powers of 2 in the set of denominators, that (⋆) is impossible?

Let ${a}_{1},{a}_{2},{b}_{1},{b}_{2},{c}_{1},{c}_{2},{d}_{1},{d}_{2},{e}_{1},{e}_{2}$ be odd integers, with

$gcd({a}_{1},{a}_{2})=gcd({b}_{1},{b}_{2})=gcd({c}_{1},{c}_{2})=gcd({d}_{1},{d}_{2})=gcd({e}_{1},{e}_{2})=1,$

and assume $n\ge 3$ and $m\ge 1$ are integers such that

$\begin{array}{}\text{(}\star \text{)}& \frac{{a}_{1}}{{2}^{3(n-1)m}{a}_{2}}+\frac{{b}_{1}}{{2}^{3((n-1)m+1)}{b}_{2}}+\frac{{c}_{1}}{{2}^{3((n-1)m+1)}{c}_{2}}+\frac{{d}_{1}}{{2}^{3nm}{d}_{2}}+\frac{{e}_{1}}{{2}^{3nm}{e}_{2}}=1.\end{array}$

Is there a way to prove, using only the powers of 2 in the set of denominators, that (⋆) is impossible?