Identifying rows and columns of a matrix that satisfy a specific feature

One possibility is:

z = Position[mat, {0 ..}];
c = Position[Transpose@mat, {0 ..}];

Delete[Transpose@Delete[mat // Transpose, c], z]

An alternative is:

Select[Transpose[Select[mat, Not[And @@ PossibleZeroQ[#]] &]], 
  Not[And @@ PossibleZeroQ[#]] &] // Transpose

nonzerorows = Flatten @ Position[mat, Except @ {0 ..}, 1, Heads -> False]

{1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15}

nonzerocols = Flatten @ Position[Transpose @ mat, Except @ {0 ..}, 1, Heads -> False]

{1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15}

mat1 = mat[[nonzerorows, nonzerocols]];

mat1 // MatrixForm

{zerorows, zerocols} = MapThread[Complement[Range @ # @ Dimensions[mat], #2] &, 
  {{First, Last}, {nonzerorows, nonzerocols}}]

 {{4, 8, 12, 16}, {4, 8, 12, 16}}

mat2 = DeleteCases[{}] @ ReplacePart[mat, 
   {Alternatives @@ zerorows, _} | {_, Alternatives @@ zerocols} -> Nothing];

mat2 == mat1

 True

You can get the final matrix directly using

mat3 = Nest[DeleteCases[{0 ..}] @* Transpose, mat, 2]

mat3 == mat1

True

A row of all zeros also sums to zero... so you can apply Total to sum the elements of each row and then select the positions that are zero:

Position[Total[mat], _?((# == 0) &)]

{{4}, {8}, {12}, {16}}

If you just want the row numbers, you can Flatten. For columns, do the same thing to the transpose of mat. If your matrix has both positive and negative entries, apply Abs to mat.