Answer: 738

Small Hint:

$c$ is irrelevant: $p(m)-p(2)=(m-2)q(m),$ where the three terms of the quadratic $q(m)$ are $m^2,$ $(a+2)m,$ and $2a+b+4.$

Big Hint:

A monic integer quadratic with one integer root has two, so its root set must be $\{2, k\}$ or a double root $k,$ with $k \ne 2.$

Solution:

Since $p(m) - p(2)$ does not involve $c,$ each valid pair $(a, b)$ contributes $41$ polynomials. Factoring, $$\begin{aligned} p(x)-p(2) &= (x-2)q(x), \\ q(x) &= x^2+(a+2)x \\ &\quad {}+(2a+b+4), \end{aligned}$$ so we need the quadratic factor $q(x)$ to have exactly one integer root different from $2.$ If $q$ has any integer root, its other root is also an integer (their sum $-(a+2)$ is an integer); if $q$ has no integer root, then no $m$ exists at all. So either $q$ has roots $2$ and $k$ with $k \ne 2,$ or a double root $k \ne 2.$

Roots $2$ and $k:$ Vieta’s formulas give $a = -4 - k$ and $b = 4k + 4.$ The constraint $-20 \le b \le 20$ forces $-6 \le k \le 4$ (and then $a$ is automatically in range), so excluding $k = 2$ leaves $10$ pairs. Double root $k:$ here $a = -2k - 2$ and $b = k^2 + 4k,$ and $b \le 20$ forces $-6 \le k \le 2,$ all valid for $a,$ so excluding $k = 2$ leaves $8$ pairs.

That is $18$ pairs $(a, b),$ hence $18 \cdot 41 = 738$ polynomials.