Here is a solution to the hint:
First of all, note that since all the ideals in question contain $\mathfrak a$, we may replace
$A$ by $A/\mathfrak a$, and so assume that $\mathfrak a = 0$; this simplifies the notation somewhat.
Next, the condition that $\mathfrak b_1 \cap \cdots \cap \mathfrak b_r = 0$ is equivalent
to the requirement that the natural map $A \to A/\mathfrak b_1 \times \cdot \times
A/\mathfrak b_r$ (the product ot the natural quotient maps) is injective, while the
condition that $\mathfrak b_i$ is irreducible is equivalent to the statement that
if $I$ and $J$ are non-zero ideals in $A/\mathfrak b_i$, then $I \cap J \neq 0$ also.
Now suppose given our two irreducible decompositions of $0$. Choose $i$ as in the hint,
and set $I_j := \mathfrak b_1 \cap \cdots \cap \mathfrak b_{i-1} \cap \mathfrak c_j\cap \mathfrak b_{i+1}
\cap \cdots \cap \mathfrak b_r,$ for each $j =1,\ldots,s$.
Then $I_1\cap \cdots \cap I_s = 0$ (since it is contained in the intersection
of the $\mathfrak c_j$, which already vanishes).
Now we recall that $A$ embeds into the product of the $A/\mathfrak b_{i'}$.
Note that $I_j$ is contained in $\mathfrak b_{i'}$ for $i'\neq i$. Thus,
if we let $J_j$ denote the image of $I_j$ in $A/\mathfrak b_i$, then
we see that the image of $I_j$ under the embedding
$A \hookrightarrow A/\mathfrak b_1\times\cdots\times A/\mathfrak b_i \times \cdots \times A/\mathfrak b_r$
is equal to $0 \times \cdots \times J_j \times\cdots \times 0$.
Thus the intersection of the images of the $I_j$, which is the image
of the intersection of the $I_j$ (since we looking at images under an embedding)
is equal to $0\times \cdots \times (\bigcap J_j) \times \cdots \times 0.$
Thus, since the intersection of the $I_j$ is equal to $0$, we see
that $\bigcap J_j = 0.$ But $\mathfrak b_i$ is irreducible, and so
one of the $J_j = 0$. Equivalently, the corresponding $I_j = 0.$
This proves the hint.
(I think the exercise should be a fairly easy deduction from the hint. The
statement that $r = s$ at least follows directly, using the hint together with
minimality of the two decompositions.)
With your definition of Noetherian, then you don't need full AC to carry out the argument, but only the weaker principle known as Dependent Choices (DC), which asserts that one can make countably many choices in succession. In your argument, if there is no maximal ideal, then by DC you could successively pick larger and larger ones, violating the Noetherian property.
Best Answer
Since you asked for a proof, let me complement Chris Phan's answer by outlining a proof that relies only on the Compactness Theorem for propositional logic, which is yet another equivalent to the Ultrafilter Theorem over ZF.
Let A be a commutative ring and let x ∉ Nil(A). To each element a ∈ A associate a propositional variable pa and let T be the theory whose axioms are
Models of T correspond precisely to prime ideals that do not contain x. Indeed, if P is such an ideal, then setting pa to be true iff a ∈ P satisfies all of the above axioms, and conversely. So it suffices to show that T has a model.
Since xn ≠ 0 for all n, one can verify using ideals over finitely generated subrings of A that the theory T is finitely consistent, i.e. any finite subset of T has a model. (What I just swept under the rug here is a constructive proof of the theorem for quotients of Z[v1,...,vn].) The Compactness Theorem for propositional logic then ensures that T has a model.