newtonian mechanics - Block on a block problem, with friction

Consider two blocks, one on top of the other on a frictionless table, with masses $m_1$ and $m_2$ respectively. There is appreciable friction between the blocks, with coefficients $\mu_s$ and $\mu_k$ for static and kinetic respectively. I'm considering the fairly routine problem of determining the maximum horizontal force $F$ (say, to the right) that can be applied to the top block so that the two blocks accelerate together.

The problem is not hard to solve symbolically. If the two blocks move together, their accelerations are the same, and the top block doesn't move with respect to the bottom block, so only static friction is in play. In a standard coordinate system (with $x$ oriented to the right), the sum of horizontal forces for the top block is

$$F-F_{sf}=m_1a$$

and for the bottom block

$$F_{sf}=m_2a$$

where $F_{sf}$ is the force of static friction. Solving for $a$ in these two expressions, and then equating them, gives

$$F=\frac{(m_1+m_2)F_{sf}}{m_2}$$

The maximum such force will therefore be achieved when $F_{sf}$ is maxed out at $\mu_s m_1g$, so

$$F_{max}=\frac{m_1}{m_2}\mu_s(m_1+m_2)g$$

I understand this solution, but conceptually I don't have a response to the following nagging question: $F_{max}$ is clearly larger than the max static friction force $\mu_sm_1g$ (because $\frac{m_1+m_2}{m_2}>1$), so why doesn't the application of a force of magnitude $F_{max}$ to the top block cause kinetic friction to take over? This line of reasoning would suggest that applying a force $F$ of magnitude greater than $\mu_sm_1g$ would cause the top block to start moving with respect to the bottom block (in which case the blocks no longer accelerate together, as in the above solution). I'm at a loss, conceptually, to say what's wrong here. I suspect it has something to do with being careful about reference frames, but a clear explanation would be much appreciated.