Is the language $L = \{a^nb^m : n = 2^m\}$ context-free?
You have stated that pumping $s$ $i$ times will give a string which contains $2^p+(i-1)*j$ $a$'s and $p+(i-1)*k$ $b$'s where $1 \le j+k \le p$. For the given language $L$ to be a CFG, new string must also belong to $L$ for all valid values of $i$. 

Put $i=2$. This gives a string containing $2^p+j$ $a$'s and $p+k$ $b$'s. For this to be of the form $a^{2^m}b^m$, we need $$2^{p+k} = 2^p+j$$ This implies $$j = 2^p*(2^k-1)$$ 

Case 1: If $k = 0$, then from the above equation, $j$ must be $0$. But, this implies, $|vy| = 0$. So, this case is not possible

Case 2: If $k \ne 0 $, then $j \ge 2^p \ge p$, but this is also not possible since $|vy| \le p$. So, the equation  $$j = 2^p*(2^k-1)$$ is not possible.

Therefore, we cannot split the string $s = a^{2^p}b^p$ satisfying all constraints. So, the language $L$ is not Context Free.