No.
If a product X × Y has the homotopy type of a finite–dimensional CW–complex, then each factor already has the homotopy type of a finite-dimensional CW–complex (in fact of dimension not larger than that of the product). Hence no counter–example can exist.

Outline of the argument

Let

  f : X × Y → K, g : K → X × Y

be homotopy inverses, with K an n–dimensional CW–complex.
Fix

  y0 ∈ Y and x0 ∈ X.

1. Produce maps between X and K

 i_X : X → X × {y0} ⊂ X × Y, 
 a = f ∘ i_X : X → K (1)

 r_X = π_X ∘ g : K → X (2)

 (π_X denotes the projection X × Y → X.)

2. Verify that r_X∘a is the identity on X (up to homotopy)

 r_X ∘ a  = π_X ∘ g ∘ f ∘ i_X  ≃ π_X ∘ id_{X×Y} ∘ i_X  = π_X ∘ i_X  = id_X. (3)

So X is a retract up to homotopy of K; we say that X is dominated by an n–dimensional CW–complex.

3. Spaces dominated by an n–complex are n–complexes
   A classical result of Whitehead (or see, e.g., Switzer, Algebraic Topology, Th. 8.11) states:

  If a space Z is dominated by an n–dimensional CW–complex,   then Z itself has the homotopy type of an n–dimensional CW–complex.

Applying this to Z = X we conclude that X ≃ X′ where X′ is an n–dimensional CW–complex.

4. Do the same for Y
   Use the inclusion j_Y : {x0} × Y → X × Y instead of i_X, and the projection π_Y : X × Y → Y, to produce maps

 a′ = f ∘ j_Y : Y → K,  r_Y = π_Y ∘ g : K → Y,

with r_Y ∘ a′ ≃ id_Y. Hence Y, too, is dominated by K and therefore has the homotopy type of an n–dimensional CW–complex.

Consequences

• Both factors inherit finite dimensionality from the product; indeed dim X, dim Y ≤ dim K.
• Because of this, there can be no space X that fails to be homotopy equivalent to a finite-dimensional CW-complex while admitting a Y with X × Y finite-dimensional up to homotopy.

Thus the answer to the original question is negative: whenever X × Y ≃ K with K finite-dimensional, both X and Y are themselves homotopy equivalent to finite-dimensional CW-complexes.