Can two different topological spaces cover each other?

I.e. do there exist non-homeomorphic $X$, $Y$ and covering maps $f:X\rightarrow Y$, $g: Y\rightarrow X$?

I have a basic understanding of covering space theory as its taught in school.

I was inspired by this question Two covering spaces covering each other are equivalent?

I have done some of the things you would do, looking at what happens to fundamental groups. And I've tried to find counter examples by drawing graphs.

The answer is yes and X and Y can even be surfaces.

Let $X_1$ be the twice punctured torus, $X_2$ the 4 punctured sphere, $Y_1$ the once punctured torus, and $Y_2$ the thrice punctured sphere.

Here is a proof that either $X_1$ and $Y_1$ mutually cover each other, or $X_2$ and $Y_2$ mutually cover each other. Note that there is a natural 2-sheeted covering from $X_1$ to $Y_1$ and likewise for $X_2$ and $Y_2$ so we only need to establish the other direction.

Now $\pi_1(X_i) = F(3)$ and $\pi_1(Y_i) = F(2)$, where $F(n)$ is the free group on $n$ generators. Note that $F(2)$ is a natural subgroup of $F(3)$ so the correspondence between subgroups of the fundamental group and covering spaces guarantees that there is a covering space of each $X_i$ which have fundamental group $F(2)$, and since $X_i$ is a surface the covering space must also be a surface.

It is a fairly straightforward exercise from an introductory course on the fundamental group to show that the fundamental group of a genus $g$ surface with $m$ punctures is $F(m+2g-1)$. A less obvious observation is that the fundamental group of a non-punctured surface is not free. Here is a link with quite a few proofs of this fact, most aren't that elementary unfortunately.

This tells us that $Y_1$ and $Y_2$ are the only surfaces which have $\pi_1 = F(2)$. This means that either $Y_1$ or $Y_2$ covers $X_1$ and the same is true with $X_2$. Obviously if either $Y_1$ covers $X_1$ or $Y_2$ covers $X_2$ we are done, since they mutually cover each other and are not homeomorphic. So assume for the sake of contradiction that neither of those coverings exist.

This implies that $Y_1$ must cover $X_2$ and $Y_2$ must cover $X_1$. But the composition of covers of $Y_1$ to $X_2$, $X_2$ to $Y_2$, and $Y_2$ to $X_1$ gives a covering map from $Y_1$ to $X_1$ which is a contradiction.

So either the once and twice punctured tori cover each other or the 3 and 4 punctured spheres do. (My money is on the spheres, but I could be wrong).

Sadly this proof isn't strictly constructive as it only gives you 2 possible pairs of spaces. If anyone has a simpler proof, or a more constructive example (or if they can rule out one of the pair, or prove that it's both) that would be awesome.