About your example: preferring the notation $\mathfrak a$ to $M$, it is easier to think of your ring $A \times \mathfrak a$ as the Rees algebra of $\mathfrak a$ over $A$, namely $R = A[\mathfrak a t] = \bigoplus_{n \ge 0} \mathfrak a^n t^n$ (the variable $t$ is used to keep track of the degree of an element), modulo the ideal of all elements of $R$ of degree $\ge 2$. Under this notation, it is clear that if $\mathfrak p \in \mathrm{Spec}(A)$, then $(\mathfrak p, \mathfrak a t) \in \mathrm{Spec}(A)$ since $R/(\mathfrak p, \mathfrak a t) \simeq A/\mathfrak p$ is an integral domain. Now let $\mathfrak m = (x_1,\cdots,x_n)_A$ be the ideal of $A$ generated by the polynomial variables, and let us show that $\mathfrak b \overset{def}= (\mathfrak m)_R$ is a counter-example (as you did). Since one easily sees that $\mathfrak a t \subseteq \mathrm{Nil}(R)$, if $\mathfrak b$ contained a prime ideal $\mathfrak q$, then it would contain $\mathfrak a t \subseteq \mathrm{Nil}(R) \subseteq \mathfrak q$ as well, so we would expect
$$
\mathfrak a t = \mathfrak b \cap \mathfrak at = \mathfrak m (\mathfrak a t) = (\mathfrak m \mathfrak a)t
$$
because the only way to get an element of $t$-degree 1 by multiplying an element of $t$-degree $0$ in $\mathfrak m$ by something is if the element of $t$-degree $0$ in $\mathfrak m$ is multiplied by an element of $t$-degree $1$, and those all belong to $\mathfrak a t$. By Nakayama's Lemma (which we can use since $\mathfrak m$ is a maximal ideal of $A$), $\mathfrak a = \mathfrak m \mathfrak a$ implies $\mathfrak a = 0$, which is assumed false. So $\mathfrak b$ contains no prime ideal.
After that, as you mentioned, any minimal prime ideal above $\mathfrak b$ must contain $\mathfrak b \oplus \mathfrak a t$ since $\mathfrak a t \subseteq \mathrm{Nil}(R)$. If $\mathfrak q$ is a prime ideal containing $\mathfrak b$, then it contains $\mathfrak m$, and it contains $\mathfrak a t$ since those elements are nilpotent, so
$$
\mathfrak q = \mathfrak m \oplus \mathfrak a t \supsetneq \mathfrak m \oplus \mathfrak m \mathfrak a t = \mathfrak b
$$
because $\mathfrak q$ is maximal (and in fact $R/\mathfrak q \simeq A/\mathfrak m$). Therefore, $\mathfrak q$ is the unique prime ideal containing $\mathfrak b$, and thus $\mathrm{ht}(\mathfrak b) = \mathrm{ht}(\mathfrak q)$. By construction, the ring extension $A \subseteq R$ is integral, and therefore $\mathrm{ht}(\mathfrak b) = \mathrm{ht}(\mathfrak q) \ge \mathrm{ht}(\mathfrak m) > 1$.
I took care to not use anywhere the explicit expression of $\mathfrak m = (x_1,\cdots,x_n)_A$ that you picked in your example where $A = F[x_1,\cdots,x_n]$; nor did I exploit the fact that $\mathfrak a^2 = 0$, I only needed that $\mathfrak a t \subseteq \mathrm{Nil}(A)$. This means that for any non-zero ideal $\mathfrak a \trianglelefteq A$ that contains no non-zero prime ideals of $A$, the ideal $\mathfrak a t \subseteq A[\mathfrak a t]$ together with any maximal ideal $\mathfrak m$ of $A$ and the corresponding ideal $\mathfrak b = (\mathfrak m)_R$ provides an example of an ideal $\mathfrak b \trianglelefteq R$ containing no non-zero prime ideals and $\mathrm{ht}(\mathfrak b) \ge \mathrm{ht}(\mathfrak m)$, as long as $\mathfrak a t$ is nilpotent in $R$; furthermore, if $A$ is integrally closed, the integral extension $A \subseteq R$ satisfies the going-down theorem and $\mathrm{ht}(\mathfrak b) = \mathrm{ht}(\mathfrak m)$.
In particular, since you didn't explicitly use the fact that $A$ was an integral domain but only the fact that $F$ was a field to have your maximal ideal of height $n$ (and help you through the proof), it suffices to pick $A$ a ring with an ideal $0 \subsetneq \mathfrak a \subseteq \mathrm{Nil}(A)$ (you can take equality iff $\mathrm{Nil}(A)$ is not prime), since such an ideal cannot contain any prime ideals by the definition of $\mathrm{Nil}(A)$ and $\mathfrak at$ automatically becomes nilpotent. If you pick $A$ integrally closed and a maximal ideal $\mathfrak m$ of height $n$ (or infinity), you can also find an example $(R,\mathfrak b)$ where $\mathfrak b \trianglelefteq R$ and $\mathrm{ht}(\mathfrak b) = n$.
As for the geometric significance of this construction, the Rees algebra $A[\mathfrak a t]$ corresponds to the blow-up of the affine scheme $\mathrm{Spec}(A)$ at the sheaf of ideals generated by $\mathfrak a$ (in algebraic geometry terms); if we restrict our attention to non-zero nilpotent ideals $\mathfrak a$ not equal to $\mathrm{Nil}(A)$, I can't really give you a geometric picture that's comfortable since the only comfortable blow-ups I know are those at a geometric point like a concrete point in $n$-dimensional affine space over a field or blowing up at some subvariety. Blowing up at a nilpotent ideal is a bit strange because it satisfies the two following properties:
- Nilpotent elements $a \in \mathfrak a$ now satisfy $(at)^2 = 0$ in $A[\mathfrak at]$
- The map $\mathfrak p \mapsto \mathfrak p \oplus \mathfrak at$ from $\mathrm{Spec}(A)$ to $\mathrm{Spec}(A[\mathfrak at])$ is a homeomorphism and comes from a closed immersion induced by the quotient map $A[\mathfrak at] \subseteq A[t] \to A$ sending $t$ to $0$. (In fact, seeing it like this, one notices that the property that $A$ is integrally closed is not really necessary in the above arguments; the homeomorphism property follows from the fact that $\mathfrak a$ is nilpotent.)
Let's look at a specific example to develop some intuition. Let $F$ be a field and consider $A = F[x_1,\cdots,x_n, x]/(x^n)$. We have $\mathrm{Nil}(A) = (x)/(x^n)$ which contains no non-zero prime ideals, so we can actually take $\mathfrak a = (x^k)/(x^n) \subseteq \mathrm{Nil}(A)$ for some $1 < k \le n$ (we can't take $k=1$ since in this case $(x^k)/(x^n)$ is prime). The Rees algebra $R$ of $\mathfrak a$ in $A$ is equal to $F[x_1,\cdots,x_n,x,x^kt]/(x^n) \subseteq F[x_1,\cdots,x_n, x,t]/(x^n)$. Clearly $x$ and $t$ are algebraically independent, so $R \simeq F[x_1,\cdots,x_n,x,u]/(x^n, u^{\lceil n/k \rceil})$ (just set $u = x^kt$). So the ring $F[x_1,\cdots,x_n,x]/(x^n)$ is like the point $0 \in \mathbb A^n_F$ that has been "fattened", and blowing-up at $(x^k)/(x^n)$ for $1 < k \le n$ is like making the point $0 \in \mathbb A^n_F$ "even fatter, but in a different direction" because you're blowing it up in this new direction $t$ while still preserving the fatness of that point you already had. But since we just fattened the already fat point, we keep the same points as before blowing up.
Our examples of ideals $\mathfrak b$ are sort of ideals representing "different degrees of fatness" of this point. They cannot contain any prime ideal $\mathfrak q$, because the inclusion $\mathfrak q \subseteq \mathfrak b$ is equivalent to the existence of a closed immersion $A/\mathfrak q \to A/\mathfrak b$, which would mean our "fat point" would contain a proper subvariety; $\mathfrak b$ is a fat point, so the only thing this ideal contains is "fatter points", and going fatter and fatter, you eventually obtain the zero ideal (the fattest point). But none of these fatter points are prime since they are all "fattened by nilpotents". However, the height of this fat point is the same as that of the corresponding point in the non-fattened variety.
Replace "geometric point in $\mathbb A^n_F$" by "some closed point of $\mathrm{Spec}(A)$" and you go back to the arbitrary setting of algebraic geometry and affine schemes. As a note, when $\mathfrak a \trianglelefteq A$ is an ideal, the associated graded ring $\mathrm{gr}_A(\mathfrak a) \overset{def}= (A/\mathfrak a)[(\mathfrak a/\mathfrak a^2)t]$ has also been extensively studied; in our context, $R/\mathfrak b = \mathrm{gr}_A(\mathfrak a) / \mathfrak m\mathrm{gr}_A(\mathfrak a)$.
Hope that helps,
