Pumping lemma with multiple of prime number + a constant
Your error is that you assume that you must find one $i$ to satisfy all the subdivisions $uvwxy$ of the lemma. The lemma is actually much weaker in that you need to find any $i$ for every subdivision $uvwxy$ of a string $s$ in $L$ longer than $h$. Without loss of generality, I will assume that any given $h$ is prime, because if it were not, I would just substitude all next occurances with any higher prime number and everything would still work the same way (I never needed the condition $|vwx|\le h$).

That said you can take your second try from the question and then fork it into two cases where:


$v \neq \epsilon$ (where $\epsilon$ means empty string, I don't know if that's international notaion. It can be otherwise written as $|v|\neq0$ )
From your first equation $$h + |v|(i-1) = m, m \text{ prime}$$ with $i = h + 1$ you get $$h + |v|h = (1+|v|)h$$ which is composite number because of our assumption $v \ne \epsilon$, meaning the string is not from $L$.
$|v|=0$ where we can pick another $i$ because it is different division of the string $s$. Here we get $$|v|=0 \land |vx| \ge 1 \implies |x|\ge 1$$
And we can trivially chose $i=0$, because now we actually subtract from the second string of zeroes, which was originally $2h+1$ long. The new length will be shorter than that, meaning we'll get a string $0^{h}10^{g}$ where $g&lt;2h+1$ thus the string is not from the language $L$.


On final note, if you were to use this as a rigorous proof, you would need to address all the other options including those, where the $1$ is part of either of the pumped strings, it is part of either $u$ or $y$ and so on. I covered only the two interesting options and the others should now be trivial to you (I assume that is why you never mentioned them in your question and assumed that 1 is part of the $w$ string).