What assumption on $x$ is needed for $x x^\top$ to be positive definite? Consider  $M = x x^\top$, $x \in \mathbb{R}^n$
For $n = 1$: $M = x^2$, then $M$ is PD if $x \neq 0$.
For $n = 2$, $M = \begin{bmatrix} x_1^2 & x_1x_2 \\ x_2x_1 & x_2^2 \end{bmatrix}$, which is PD when $x_1 \neq x_2$ and $x_1 \neq 0 \wedge x_2 \neq 0$
Is there a pattern to the positive definiteness or lack thereof of this matrix? 

Observe that for any vector $v\in\mathbb R^n$ we have that
$$
v^TMv=(v^Tx)(x^Tv)=\langle v,x\rangle^2.
$$
For $M$ to be positive definite, we need this to be strictly greater than zero for all non-zero $v$. But this is impossible if $n> 1$, since the orthogonal complement of the subspace spanned by $x$ is non-empty.
In particular, your $n=2$ is never positive definite since
$$
M\begin{pmatrix}x_2\\-x_1\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}.
$$
Another explanation:
The matrix $xx^T$ has rank at most $1$ (by the rank inequality mentioned here), which implies that it has determinant zero when $n>1$ (since non-zero determinant implies that the rank equals $n$). And a positive definite matrix cannot have determinant zero.
A third explanation:
By applying a linear transformation to $x$ such that it becomes a standard basis vector, you obtain a similarity transformation of $xx^T$ into a matrix that has at most one non-zero entry.
