Let $R$ be a commutative ring. If $I\leq R$ is an ideal s.t. $\sqrt I=m$ is maximal, then $I$ is primary.
Proof
Since $m$ maximal, $Nil(R/I)=m/I=:n$ is maximal.
Q1) Why $m/I$ is maximal in $R/I$ when $m$ is maximal in $R$ ? I have problem to prove this fact.
Hence, every prime ideal contain $n$.
Q2) Why is $p$ is prime in $R/I$, then $p\supset n$ ?
1)
The quotient of $R/I$ by $m/I$ is $R/m$ by the third isomorphism theorem. It is a field, so that $m/I$ is a maximal ideal of $R/I$.
Alternatively, any ideal of $R/I$ is of the form $J/I$, where $J$ is an ideal of $R$ containing $I$. So if $J/I$ contains $m/I$, can you prove that $J/I=m/I$ or $J/I=R/I$?
2)
You know that $$\mathrm{Nil}(R/I) = \{x \in R/I \mid \exists r>0, x^r=0\}.$$ So proving that any prime of $R/I$ contains $n$ is equivalent to proving that $$n \subset \bigcap_{p \; \text{prime}} p = \mathrm{Nil}(R/I).$$
Assume that $x \in n = \sqrt I / I$. Then $x^r = 0$ for some $r>0$, since $x=[y]_{I}$ for some $y \in \sqrt I$. This shows that $x \in \mathrm{Nil}(R/I)$, hence $n \subset p$ for any prime ideal $p$ of $R/I$.
Therefore $R/I$ has only $n=m/I$ as prime ideal. If $x \in R/I$ is a zero divisor, then it is not a unit, then it belongs to $n$ (because any proper ideal, for instance $(x) \neq R$, is contained in a maximal ideal — but here $n$ is the only maximal ideal of $R/I$). Then $x=[y]_I$ where $y \in \sqrt I$, which implies that $x$ is nilpotent.
Hence, $I$ is $m$-primary.