Show $\{1^n0^m |\space n \neq 2^m\}$ not regular using pumping lemma

Showing that the language $ L$ with $ \{1^n0^m |\space n \neq 2^m\}$ is not regular using Myhill-Nerode is easy: Let $ i, j\in \mathbb{N}.i\neq j.$ It follows $ 1^{2^i}\nsim 1^{2^j}$ because $ 1^{2^i}0^{i}\notin L$ but $ 1^{2^j}0^{i}\in L$ . Therefore $ L$ has an infinite amount of Myhill-Nerode equivalence classes and is not regular. But how do I show this using the general version of the pumping lemma for regular languages?