Let the quotient of the division be x^11 + Jx^10 + ... + Bx^2 + Ax + Z. All the coefficients must be integers.
Multiplying the quotient by x^2 - x + N yields x^13 + (J-1)x^2 + ... + (B*N-A+Z)x^2 + (A*N-Z)x + Z*N. This must equal x^13 + x + 90.
From the constant term: Z*N = 90 implies Z = 90/N, therefore N is a factor of 90. There are 24: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90, -1, -2, -3, -5, -6, -9, -10, -15, -18, -30, -45, -90.
From the linear term A*N-Z = 1 implies A = (Z+1)/N. There are six values of N which have N, Z, and A integral: (N,Z,A) = (1,90,91), (2,45,23), (10,9,1), (-1,-90,89), (-2,-45,22), (-9,-10,1).
From the quadratic term: B*N-A+Z = 0 implies B = (A-Z)/N. Only three of the six possible candidates make B an integer: (N,Z,A,B) = (1,90,91,1), (2,45,23,-11), (-1,-90,89,-179)
With only three possible N, trial division is feasible. Of N=1, -1, and 2 the division works for only N=2. Specifically x^13 + x + 90 = (x^2 - x + 2) * (x^11 + x^10 - x^9 - 3x^8 - x^7 + 5x^6 + 7x^5 - 3x^4 - 17x^3 - 11x^2 + 23x + 45).
Solution