Solution:

There do not exist such positive integers. Assume for a contradiction that there do, however. We may assume $m > n$, so that $b > a$. Since we have equality between a power of $a$ and a power of $b$, $b$ is a rational power of $a$. We write $b = a^{x}$, where we know that $x \geq 1$ is a rational number. Then we may rewrite the $b$ power tower (using $n \geq 2$) as
$$\underbrace{b^{b^{\cdots}}}_{n\ b\ \text{'s}} = \underbrace{a^{x a^{x b} \cdots^{\cdot b}}}_{n-2\ b\ \text{'s}}$$
On the other hand, we know that this is equal to the power tower of $a$'s in the given equation, so removing the bottom $a$ gives
$$\underbrace{a^{a^{\cdots a}}}_{m-1\ a'\mathrm{s}} = \underbrace{x a^{x b^{\cdots b}}}_{n-2\ b\ b'\mathrm{s}}$$
In particular, this tells us that $x$ is a rational power of $a$, $x = a^{y}$ for some nonnegative rational number $y$. Substituting and again removing the bottom $a$ gives
$$\underbrace{a^{a^{\cdots}}}_{m-2} = y + a^{y} \underbrace{b^{b^{\cdots}}}_{n-2}$$
Observe that the right side is $y$ plus a rational power of $a$ (simply $a^{y}$ if $n=2$). Now consider the possibilities for this power $a^{p/q}$. If $p/q < 0$, then $y < 0$ and the right side of (2) is less than 1, an impossibility. So either $a^{p/q}$ is irrational (again impossible) or it is the $q$th root of a $q$th power, which is necessarily a positive integer. Then $y$ is also an integer, namely, the difference between the tower of $m-2$ $a$'s on the left and another power of $a$. Dividing both sides of (2) by $y$ indicates that $y$ is at least $1 - \frac{1}{a} = \frac{a-1}{a}$ times the tower of $m-2$ $a$'s, which is certainly larger than a power tower of $m-3$ $a$'s. But then $a^{y}$ is larger than a power tower of $m-2$ $a$'s, and clearly the right side of (2) is even larger than this, so we have a contradiction.