What is known about Lie groups with (strictly) positive curvature?

The standard reference is:

Curvatures of left invariant metrics on lie groups by J Milnor - ‎1976

In particular, a theorem of Wallace (mentioned in Milnor's paper) confirms your conjecture.

(PS: The paper seems to be available to all).

The following result is, for example, exercise 3 on pg. 104 of Do Carmo's Riemannian Geometry book.

Suppose $X$ is a Killing field on a compact even dimensional Riemannian manifold of positive curvature. Then $X$ has a zero.

Using this result, it's very easy to prove the following generalization of Wallach's theorem.

Theorem: No compact Lie group $G$ of rank $2$ or higher admits a positively curved Riemannian metric which is invariant under left multiplication by any $T^2\subseteq G$.

Proof: Suppose there is such a group $G$. Because the $T^2$ action on $G$ by left multiplication is free, the action fields associated to it have no zeros. On the other hand, because the $T^2$ is isometric, the action fields are are Killing fields. By the above result, this implies the dimension of $G$ is odd.

Because the action of $S^1\times \{1\}\subseteq T^2$ on $G$ is isometric, the even dimensional manifold $G/(S^1\times \{1\})$ inherits a Riemannian metric for which the action by $\{1\}\times S^1$ is free and isometric. Further, by O'Neill's formulas for a Riemannian submersion, the induced metric on $G/(S^1\times \{1\})$ is positively curved. But then, just as above, the action field is non-zero and Killing, contradicting the quoted result above. $\square$

Incidentally, without some kind of symmetry assumption, we can say basically nothing about the existence of positively curved metrics on compact Lie groups with finite $\pi_1$. The issue that among closed simply connected manifolds, there is no known obstruction which distinguishes those that admits metrics of non-negative curvature from those which admit metrics of positive curvature.