Question: "I wanted to see some applications of "going up" and/or "going down" to compute the prime & maximal ideals of $\mathbb{Z}[\sqrt{D}]$ whenever $D$ is square-free."
Answer: Note: If $K:=\mathbb{Q}[t]/(t^2-D)$ with $D$ a square free integer and $D\cong2,3(mod 4)$ it follows the ring of integers $\mathcal{O}_K$ satifies
$$A:=\mathcal{O}_K \cong \mathbb{Z}[t]/(t^2-D).$$
This is also written as $\mathcal{O}_K=\mathbb{Z}[\sqrt{D}]$ (it is not clear what you mean when you write $\mathbb{Z}(\sqrt{D})$, as mentioned in the comments).
The extension $R:=\mathbb{Z} \subseteq A$ is integral hence for any prime ideal $(p) \subseteq \mathbb{Z}$ there is a prime ideal $I \subseteq A$ with $I\cap R =(p)$.
Given a prime ideal $(p) \subseteq R$ you get an ideal $(p)A \subseteq A$ and a quotient $A/(p)A \cong k[t]/(t^2-\overline{D})$ with $k:=\mathbb{Z}/(p)$ and $A/(p)A$ is a field iff the polynomial $p(t):=t^2-\overline{D}$ is irreducible in $k[t]$. The polynomial $p(t)$ is reducible iff the equation $t^2\cong \overline{D}(mod (p))$ has a solution in $k$. There is a classical result in algebraic number theory giving an algorithm to determine this: The "Gauss reciprocity law" (Neukirch, Algebraic number theory Thm.8.6).
Example: If $A:=\mathbb{Z}[i]$ is the ring of Gaussian integers you may use this principle to classify prime ideals in $A$. Since $A$ is a PID it follows any prime ideal $I:=(a+ib)$ is generated by an element $a+ib$ with $a,b \in \mathbb{Z}$. If $(a,b)=1$, there is an isomorphism of rings
$$\mathbb{Z}[i]/(a+ib) \cong \mathbb{Z}/(a^2+b^2),$$
and it follows $(a+ib)$ is a prime ideal iff $a^2+b^2=p$ is a prime number. Such a prime number must satisfy $p \cong 1(mod (4))$. If $(p)$ is an odd prime with $p \cong 2,3(mod(4))$ it follows $(p)A$ is a prime ideal in $A$.
If $p=2$ it follows $(2)A=(1+i)^2$, hence $(2)A$ is not a prime ideal in $A$.
Hence the extended ideal $(p)A$ is prime iff $p\cong 2,3(mod(4))$.
Question: "I was wondering about the following 1-What are the prime ideals of Z((√D) whenever D is square-free."
Answer: In the case of $A:=\mathbb{Z}[i]$ you get the following classification: A prime ideal $I:=(a+ib) \subseteq A$ is generated by one element. If $b \neq 0 $ it follows $(a,b)=1$ hence
$$A/I \cong \mathbb{Z}/(p):=\mathbb{F}_p$$
for $p=a^2+b^2$ and $p$ an odd prime with $p\cong 1(mod(4))$.
If $b=0$ it follows $I=(p)A$ with $p$ a prime number. It follows $p\cong 2,3(mod(4))$. The residue field in case one is $\mathbb{F}_p$ in case two it is $\mathbb{F}_{p^2}$.
The ideal $I=(1+i)$ is prime with residue field $\mathbb{F}_2$
