Since this is a problem in polynomials, we will assume
that m is a positive integer.

Dividend = x^m+nx
Divisor = x^2-x-2
Remainder = 2x+6
Quotient = ?? = Q(x)

Dividend = Divisor∙Quotient + Remainder 

x^m+nx = (x^2-x-2)Q(x) + 2x+6

x^m+nx = (x-2)(x+1)Q(x) + 2x+6

Substitute x=2

      2^m+n∙2 = (2-2)(2+1)Q(2) + 2∙2+6
      2^m+2n = 0 + 4+6
(1)   2^m+2n = 10


Substitute x=-1

      (-1)^m+n∙(-1) = (-1-2)(-1+1)Q(-1) + 2∙(-1)+6
      (-1)^m-n = 0 + 2∙(-1)+6
      (-1)^m-n = -2+6
      (-1)^m-n = 4
            -n = 4 - (-1)^m
(2)          n = -4 + (-1)^m

m is either even or odd.  Assume m is even, then
we would have

             n = -4 + 1
             n = -3

Substitute in equation (1)

        2^m+2n = 10
     2^m+2(-3) = 10
         2^m-6 = 10
           2^m = 16
           2^m = 2^4
             m = 4

That's one solution m=4 and n=-3

We see if there is a solution assuming m is odd. 

Then (2), which is

             n = -4 + (-1)^m

then we would have

             n = -4 - 1
             n = -5

Substitute in equation (1)

        2^m+2n = 10
     2^m+2(-5) = 10
        2^m-10 = 10
           2^m = 20

There is no positive integer solution for m

So the only solution is m=4 and n=-3

Checking by long division:

                  x^2+ x+3
x^2-x-2)x^4+0x^3+0x^2-3x+0
        x^4- x^3-2x^2
             x^3+2x^2-3x
             x^3- x^2-2x
                 3x^2- x+0
                 3x^2-3x-6
                      2x+6 

Edwin