Find the 'a' such that $x^{13}+x+90=(x^2-x+a)(Q(x))$

Substituting $x = 1 \implies a \mid 92$

Substituting $x = -1 \implies (a+2)\mid 88$

Divisors of $92 = 1,2,4,23,46,92$

Divisors of $88 = 1,2,4,8,11,22,44,88$

Thus, $a = 2$.

Indeed, checks out with Wolfram :

Inspired by Ojas's answer (which I think is incomplete because it seems to assume $a$ is positive):

$x=0\implies a\mid90$

$x=1\implies a\mid92$

Thus $a\mid2$, so $a\in\{\pm1,\pm2\}$

$x=-1\implies(a+2)\mid88\implies a\in\{-1,2\}$

Finally, we can rule out $a=-1$ because $x^2-x-1$ has a root in $[-1,0]$ while $x^{13}+x+90$ clearly does not. Thus $a=2$ is the only remaining possibility.

Remark: This doesn't prove that $x^2-x+2$ necessarily is a factor of $x^{13}+x+90$. All it proves is the following: If $a$ is an integer such that $x^2-x+a$ divides $x^{13}+x+90$, then $a=2$.