Theorem: Let $J\subseteq A$ be a maximal ideal. Then $J$ is hereditary (if $a\in A_+$ satisfies $a\leq b$ for some $b\in J_+$, then $a\in J_+$), strongly invariant (if $x^*x\in J_+$ then $xx^*\in J_+$) and if $a\in J_+$ then $a^t\in J_+$ for every $t>0$.
We need some notation and some lemmas first:
We define the relation $\sim$ on $A_+$ by setting $a\sim b$ if there exists $x\in A$ such that $a=xx^*$ and $b=x^*x$.
Lemma 1: If $a\leq b\sim b'$, then there exists $a'$ such that $a\sim a'\leq b'$.
Proof: Let $x\in A$ such that $b=x^*x$ and $b'=xx^*$.
Let $x=v|x|$ be the polar decomposition in $A^{**}$.
Set $y:=va^{1/2}$ and $a':=yy^*$.
Then $y$ belongs to $A$ and we have $a=y^*y$ and $a'=yy^*=vav^*\leq vbv^*=b'$.
Lemma 2: Let $a,b,c\in A_+$ and $t>1$ such that $a\sim b\leq c^t$.
Then there exists $y\in A$ such that $a=ycy^*$.
Proof: Choose $x\in A$ such that $a=xx^*$ and $b=x^*x$.
Set $\alpha:=\tfrac{1}{2t}$.
Then $0<\alpha<1/2$.
Applying the polar decomposition in C*-algebras (see Proposition~II.3.2.1 in Blackadar's Operator algebras book) we obtain $y\in A$ such that $x = y(c^t)^\alpha$.
Then
$$ a = xx^* = yc^{2t\alpha}y^* = ycy^*. $$
Proof of Theorem:
We consider the ideal as suggested by @Black
$$ K := \{ a\in A : a^*a \leq b \text{ for some } b\in J_+ \}. $$
We will show that $J=K$.
Indeed, as noted by @Black, we either have $J=K$ or $K=A$.
We show that $K=A$ leads to a contradiction.
So assume that $K=A$.
Let $a\in A_+$.
Since $a^{1/4}\in K$, we obtain $b\in J_+$ such that $a^{1/2}\leq b$.
Then
$$ a = a^{1/4}a^{1/2}a^{1/4} \leq a^{1/4}ba^{1/4} \sim b^{1/2}a^{1/2}b^{1/2} \leq b^2. $$
By Lemma 1, we obtain $a'$ such that $a\sim a'\leq b^2$.
It then follows from Lemma 2 that $a\in J_+$.
Thus $J=A$, a contradiction.
With an argument as in the answer of @Black, it now follows that $J$ is hereditary, strongly invariant, and closed under roots of positive elements.
