If $k$ is negative or zero, then the quadratic has two real roots. But we can easily check that the other polynomial has derivative everywhere positive and hence only one real root.
So $k$ must be positive.
If $x^2-x+k$ divides $x^{13}+x+90$, then $x^{13}+x+90=f(x)(x^2-x+k)$, where $f(x)$ is a polynomial with integer coefficients.
Let $x=0$, we see that $k$ must divide $90$. Let $x=1$, we see that it must divide 92. Hence it must divide 92-90=2. So the only possibilities are 1 and 2. Suppose $k=1$, then putting $x=2$, we have that $3$ divides $2^{13}+92$ but $2^{\text{odd}}$ is congruent to 2 mod 3, so $2^{13}+92$ is congruent to 1 mod 3. So $k$ cannot be 1.
To see that $k=2$ is possible, we write
$(x^2-x+2)(x^{11}+x^{10}-x^9-3x^8-x^7+5x^6+7x^5-3x^4-17x^3-11x^2+23x+45)=x^{13}+x+90$.