Consider the following famous problem:
Two opposite corner cells of a chessboard are removed. Prove that it is impossible to cover remaining cells with dominoes $2\times1$.
The standard proof uses coloring: the removed cells are of the same color and any domino covers 1 black and 1 white cell.
However, there is another proof via geometric group theory: for each cell write $a, b, a^{-1}, b^{-1}$ on its left, upper, right, lower border correspondingly. Suppose that chessboard without two opposite corners can be covered by dominoes. That would mean that any two elements $a, b$ of any group satisfying $a^2 b a^{-2} b^{-1}=1$ and $ab^2a^{-1}b^{-2} = 1$ will also satisfy $a^7 b a b^7 a^{-7} b^{-1} a^{-1} b^{-7}=1$. The idea here is that we go clockwise along the border of some subset of cells and intepret the written word as a relation.
When two regions have some part of border in common, it cancels out to give a new relation. (See the picture: if $xy=1$ and $y^{-1}z=1$, then $xz=1$).
So, it remains to show that $a^7 b a b^7 a^{-7} b^{-1} a^{-1} b^{-7}=1$ does not follow from $a^2 b a^{-2} b^{-1}=1$ and $ab^2a^{-1}b^{-2} = 1$. Consider $S_3$ and let $a=(1 2), b= (13)$, then $a^2=1, b^2=1$, but $abab$ and $baba$ are different, since they are distinct 3-cycles.
These two methods (colorings and group theory) can be applied to any problem of the same sort (covering some cell figure with given figures).
The question is, are there examples of problems for which one of the methods is more powerful than the other, i.e. one method gives a solution while no solution using the other method is known?
For example, there is the following problem:
A table $a\times b$ can be covered by rectangles of the form $n\times 1$ and $1 \times n$ only when $n|a$ or $n|b$.
But this problem can be solved by either of the two methods.