= Solution
Let $A=\mathbb C[x,y,z]$. This <polynomial ring> is a <unique factorization domain> of <Krull dimension> three. For a <height-one prime in a unique factorization domain> $\mathfrak p$, choose a nonzero element of it and an irreducible factor $q$ belonging to $\mathfrak p$. In a <unique factorization domain>, $q$ is prime, so $(q)$ is a nonzero <prime ideal> contained in $\mathfrak p$. Strict containment would give a chain $(0)\subsetneq(q)\subsetneq\mathfrak p$ of length two. Thus
$$
\boxed{\operatorname{ht}\mathfrak p=1\Longrightarrow\mathfrak p=(q)}.
$$
If $\mathfrak p$ has height three, it is maximal: any strictly larger proper prime would extend a length-three chain and contradict $\dim A=3$. The <Weak Hilbert Nullstellensatz> over the <algebraically closed field> $\mathbb C$ then gives
$$
\boxed{\operatorname{ht}\mathfrak p=3\Longrightarrow\mathfrak p=(x-a,y-b,z-c)}
$$
for some $a,b,c\in\mathbb C$. This proves both requested generator assertions.
For the <monomial curve with exponents three, four and five>, put
$$
g_1=y^2-xz,\qquad g_2=yz-x^3,\qquad g_3=z^2-x^2y,\qquad J=(g_1,g_2,g_3).
$$
All three map to zero under the given parametrization, so $J\subseteq I=\ker\varphi$. Reduce any <monomial> using $y^2\mapsto xz$, $yz\mapsto x^3$ and $z^2\mapsto x^2y$. Each reduction decreases the sum of the exponents of $y,z$, so it terminates. Every polynomial has, modulo $J$, the form
$$
A_0(x)+A_1(x)y+A_2(x)z.
$$
Its image is $A_0(t^3)+t^4A_1(t^3)+t^5A_2(t^3)$. These three terms occupy different exponent classes modulo three, so they cannot cancel. A zero image forces each $A_i=0$. Consequently
$$
\boxed{I=(y^2-xz,\ yz-x^3,\ z^2-x^2y)}.
$$
The quotient is the domain $\mathbb C[t^3,t^4,t^5]$, hence $I$ is a <prime ideal>. It is finite free over $\mathbb C[t^3]$, with <basis of a module> $1,t^4,t^5$, by the same normal-form argument. Its dimension is therefore one, using that integral extensions preserve <Krull dimension>. The <polynomial-ring height and dimension formula> gives
$$
\boxed{\operatorname{ht}I=3-\dim(A/I)=2}.
$$
For the generator obstruction, let $\mathfrak m=(x,y,z)$. Since $I\subseteq\mathfrak m^2$, we have $\mathfrak mI\subseteq\mathfrak m^3$. The degree-two initial forms of $g_1,g_2,g_3$ are
$$
y^2-xz,\qquad yz,\qquad z^2.
$$
They are <linearly independent> over $\mathbb C$ in $\mathfrak m^2/\mathfrak m^3$. Thus no nonzero constant linear combination of the $g_i$ belongs to $\mathfrak mI$, and their classes are independent in $I/\mathfrak mI$. Since the three generators span that quotient,
$$
\boxed{\dim_{\mathbb C}I/\mathfrak mI=3}.
$$
Two global generators would span it by two vectors, which is impossible. This is the <initial-form lower bound for ideal generators>; it shows why height two does not force two generators, even though height one and height three behave as above.
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