Consider an orthonormal basis $\{x_j\}$ of $H$ with $x_1=x$, and consider $\{E_{kj}\}$ the corresponding matrix units (we don't really need matrix units, just the projection onto the span of $x$, but it might help understand). Assume that you can write $F_x=\alpha f + (1-\alpha)g$, for some $\alpha\in[0,1]$ and states $f,g$. Then, since $0\leq E_{11}\leq I$,
$$
1=\langle E_{11}x,x\rangle = F_x(E_{11})=\alpha f(E_{11}) + (1-\alpha) g(E_{11})\leq
\alpha+1-\alpha = 1.
$$
So $\alpha f(E_{11}) + (1-\alpha) g(E_{11})=1$. But as $f(E_{11})\leq1$, $g(E_{11})\leq1$, we conclude that $f(E_{11})=g(E_{11})=1$. In particular, $f(I-E_{11})=0$. Then, for any $T\in B(H)$,
$$
0\leq|f(T(I-E_{11}))|\leq f(T^*T)^{1/2} f((I-E_{11})^2)^{1/2}=f(T^*T)^{1/2}f(I-E_{11})^{1/2}=0.
$$
Thus $f(T)=f(T\,E_{11})$ for all $T$. Taking adjoints, $f(T)=f(E_{11}T)$. But then $f(T)=f(TE_{11})=f(E_{11}TE_{11})$. As $E_{11}TE_{11}=\langle Tx,x\rangle\,E_{11}=F_x(T)E_{11}$,
$$
f(T)=f(E_{11}TE_{11})=F_x(T)\,f(E_{11})=F_x(T).
$$
Similarly with $g$, so $F_x$ is an extreme point.
Edit: thanks to Matthew for noting that my argument in the last paragraph from the previous version was wrong. Here is the new argument.
Let $\pi_0:B(H)\to B(H)/K(H)$ be the quotient map onto the Calkin algebra $C(H)$. Let $\pi_1:C(H)\to B(K)$ be an irreducible representation of $C(H)$. Then
$$
\pi=\pi_1\circ\pi_0:B(H)\to B(K)
$$
is an irreducible representation. Using the correspondence between irreps and pure states, there exists a pure state $\varphi$ on $B(H)$ such that its GNS representation is unitarily equivalent to $\pi$. But this tells us that $\varphi(T)=0$ for all $T\in K(H)$, and so $\varphi$ cannot be a point state.
If $\pi :A\to B(H)$ is any non-degenerate representation of the C*-algebra $A$, and if $x$ is a unit vector in $H$, then the state
$$
\varphi (a) = \langle x,\pi(a)x\rangle
$$
is pure iff the cyclic space
$$
[\pi (A)x] = \overline{\text{span}}\{\pi (a) x: a\in A\}
$$
is irreducible for the action of $A$ there.
This is because the representation of $A$ on $[\pi (A)x]$ is equivalent to the GNS representation associated to $\varphi $, and
it is well known that a GNS representation is irreducible iff the state is pure.
In the example given, one can show that $[Ax]=\mathbb C^3$, which is not irreducible, so $\rho _x$ is not pure.
On the other hand,
as noted by @MaoWao,
if $A$ is an irreducible algebra of operators on $H$, such as $B(H)$ itself, $[\pi (A)x] = H$, for every unit vector
$x$, whence every vector state is pure.
Best Answer
This is a very nice question. Believe it or not, the result follows from Kadison's transitivity theorem (I'm not sure if someone can avoid it).
Proof:
The converse is easy and no big theorem is used, so I am leaving it as an exercise. Also, observe that the inclusion $N_\varphi+N_\varphi^*\subset\ker(\varphi)$ is true for any state $\varphi$. So the direct implication (which is what you are interested in) is about showing the other inclusion and it goes as follows.
Let $\varphi$ be a pure state and take the associated GNS representation $(H_\varphi,\pi_\varphi,\xi_\varphi)$, where $\xi_\varphi$ is the unit cyclic vector. Let $a\in\ker(\varphi)$. By the properties of the GNS construction we have that $$0=\varphi(a)=\langle\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle $$ so $\pi_\varphi(a)\xi_\varphi$ and $\xi_\varphi$ are two orthogonal vectors of $\mathcal{H}_\varphi$. Since $\varphi$ is pure, $\pi_\varphi(A)$ acts irreducibly on $\mathcal{H}_\varphi$, so we can apply Kadison's transitivity theorem: the vecotrs $\pi_\varphi(a)\xi_\varphi$ and $\xi_\varphi$ are linearly independent and the orthogonal projection $p\in\mathbb{B}(H_\varphi)$ onto the one-dimensional subspace $\text{span}\{\pi_\varphi(a)\xi_\varphi\}$ is a self-adjoint operator that maps $\pi_\varphi(a)\xi_\varphi$ to itself and it maps $\xi_\varphi$ to $0$. By Kadison's theorem, we can find a self-adjoint element $b\in A$ so that $$\pi_\varphi(b)\big(\pi_\varphi(a)\xi_\varphi\big)=\pi_\varphi(a)\xi_\varphi\text{ and }\pi_\varphi(b)\big(\xi_\varphi\big)=0.$$
Now $b\in N_\varphi^*$: indeed, $\varphi(bb^*)=\varphi(b^2)=\langle\pi_\varphi(b^2)\xi_\varphi,\xi_\varphi\rangle=\|\pi_\varphi(b)\xi_\varphi\|^2=0$. Since $N_\varphi$ is a left ideal, $N_\varphi^*$ is a right ideal, thus $ba\in N_\varphi^*$. Now $$a=(a-ba)+ba.$$ To conclude the proof, one observes that $a-ba\in N_\varphi$: $$\varphi((a-ba)^*(a-ba))=\varphi((a^*-a^*b)(a-ba))=\varphi(a^*a-2a^*ba+a^*b^2a)= $$ $$=\langle\pi_\varphi(a^*a)\xi_\varphi-2\pi_\varphi(a^*)\pi_\varphi(b)\pi_\varphi(a)\xi_\varphi+\pi_\varphi(a^*)\pi_\varphi(b)^2\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle=$$ $$\langle\pi_\varphi(a)^*\pi_\varphi(a)\xi_\varphi-2\pi_\varphi(a^*)\pi_\varphi(a)\xi_\varphi+\pi_\varphi(a^*)\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle=\langle0,\xi_\varphi\rangle=0.$$
References: I have seen this proposition in Ilijas Farah's book Combinatorial Set Theory of C$^*$-algebras. The idea of the proof is illustrated there.