$$\mathbb{Z}_{(p)}=\left\{\frac{a}{b}\mid a,b\in\mathbb{Z},p\nmid b\right\}.$$
I want to show
the maximal ideal of $\mathbb{Z}_{(p)}[x]$ which contains ideal $I=(x^2+1)$ contains $p$.
I found an ideal $(p,x^2+1)$. This is maximal and contains $I$, certainly, contains $p$.
But I don't know how to show every maximal ideal which contains $I$ contain $p$. Thank you.
Any maximal ideal containing $I$ will be the preimage of a maximal ideal $J\subset\mathbb{Z}_p[i]$. As $\mathbb{Z}_p[i]$ is not a field we know that $J\neq 0$. Pick $\frac{a+ib}c\neq 0$ in $J$. Then as $J$ is a proper ideal we have $p|a^2+b^2$. By unique factorisation in $\mathbb{Z}[i]$ we may write $$a+ib=m(u+iv)^s(u-iv)^t\qquad {\rm or}\qquad a+ib=mp^r,$$ where $m\in \mathbb{Z}[i]$ and $p\not\!|m\bar{m}$ and in the first case $u^2+v^2=p$ with $u,v\in \mathbb{Z}$.
Thus either $(u+iv)^s(u-iv)^t\in J \quad {\rm or}\quad p^r\in J$. As $J$ is prime we have either $u+iv\in J$ or $u-iv\in J$ or $p\in J$. In all three cases $p\in J$.
Alternative proof which does not use unique factorisation in $\mathbb{Z}[i]$:
Any maximal ideal containing $I$ will be the preimage of a maximal ideal $J\subset\mathbb{Z}_p[i]$. As $\mathbb{Z}_p[i]$ is not a field we know that $J\neq 0$. Pick $\frac{a+ib}c\neq 0$ in $J$, with $a,b,c\in \mathbb{Z}$. Then $$\left(\frac{a+ib}c\right)\left(\frac{a-ib}c\right)=\left(\frac{a^2+b^2}{c^2}\right)\in J.$$
Write $a^2+b^2=mp^r$ for some $m\in \mathbb{Z}$ with $p\not\!|m$ and $r\geq 0$. Then $\frac m{c^2} p^r \in J$, which is maximal hence prime. Thus either $\frac m{c^2}\in J$ or $p\in J$.
However $\frac m{c^2}$ is invertible, so not in $J$, as $J$ is a proper ideal. Thus $p\in J$.