Is there an example of a power of prime ideal in a polynomial ring that is not primary?

The question may have been addressed in Math SE before; but my question is a little variation of it, coming because of a very common example for this question.

It is well known that powers of a prime ideal need not be primary and a standard example is to consider the quotient of $k[x,y,z]$ by the ideal $(xy-z^2)$, and in the quotient ring, take the prime ideal $(\bar{x},\bar{z})$; its square is not primary.

Now this example of prime ideal is in quotient of polynomial ring.

Q. Can we construct an example of a prime ideal within a polynomial ring, say $k[x,y]$ or $k[x,y,z]$, whose some power is not primary?

The following example is taken from Northcott, Ideal Theory.

Let $R=k[x,y,z]$ and $\mathfrak{p}=(f_1,f_2,f_3)$, where $f_1=y^2-xz$, $f_2=yz-x^3$, $f_3=z^2-x^2y$.
To see that $\mathfrak{p}$ is prime, consider $R/\mathfrak{p}$ as a $k[z]$-module. Since in this quotient we have the relations $y^2=xz, yz=x^3,z^2=x^2y$, a generating system for $R/\mathfrak{p}$ as a $k[z]$-module is given by $\{1,x,y,xy,x^2\}$. It follows that any $f \in k[x,y,z]$ can be written as $$f(x,y,z)=x^2A(z)+xyB(z)+xC(z)+yD(z)+E(z)+g(x,y,z),$$ where $A,B,C,D,E \in k[z]$ and $g(x,y,z) \in \mathfrak{p}$.
Now consider the ring homomorphism $\varphi:k[x,y,z] \to k[t], f(x,y,z) \mapsto f(t^3,t^4,t^5)$. We get that $\mathfrak{p} \subseteq \operatorname{ker}(\varphi)$. If $\varphi(f)=0$, then writing $f$ in the form as above, we get $$f(t^3,t^4,t^5)=t^6A(t^5)+t^7B(t^5)+t^3C(t^5)+t^4D(t^5)+E(t^5)=0.$$ This implies that $A=B=C=D=E=0$, because the degree $d$ of every term in $t^6A(t^5)$ satisfies $d \equiv 1 \pmod{5}$, terms in $t^7B(t^5)$ have degree $d$ with $d \equiv 2 \pmod{5}$, terms in $t^3C(t^5)$ have degree $d$ with $ d \equiv 3 \pmod{5}$ etc., so there can't be any cancellation. This shows that $f(x,y,z)=g(x,y,z) \in \mathfrak{p}$. So $\mathfrak{p}=\operatorname{ker}(\varphi)$ is a prime ideal.

Now, we show that $\mathfrak{p}^2$ is not primary. If $\mathfrak{p}^2$ was primary, it would be $\mathfrak{p}$-primary. $\mathfrak{p}^2$ contains $$f_2^2-f_1f_3=(yz-x^3)^2-(y^2-xz)(z^2-x^2y)=x (x^5 - 3 x^2 y z + x y^3 + z^3),$$ as $x$ is not in $\mathfrak{p}$, so if $\mathfrak{p}^2$ was $\mathfrak{p}$-primary it would contain $x^5-3x^2yz+xy^3+z^3$, but every polynomial in $\mathfrak{p}^2=(y^2-xz,yz-x^3,z^2-x^2y)^2$ contains no terms of degree less than $4$, so this is not the case.