This is old but since there is no solution, I will add mine.
It is easy to check that all numbers below 35 satisfying the condition have solution.
Consider now $ |2^a - 3^b| = 35$.
The first case is that of $ 2^a - 3^b = 35$. It is easy to see that if a<4, there is no solution - now take modulo 16, and note that $ - 3^b = = 3 \pmod{16}$ has no solution.
The second case is $ 3^b - 2^a = 35$. By mod 4, b is odd, and by mod 3, a is even. Thus, we have an equation of the form $ 3x^2 - y^2 = 35$. Consider this modulo 7. Note that 3 is not a quadratic residue, so we are done.