Let $P(x)=-(ax^3+bx^2+cx+d)$, where the negative sign out front allows us to reformulate the condition as one of the integer coefficients $a$, $b$, $c$, or $d$ is at least $2$. Now suppose the condition is not met, so $a,b,c,d\le1$.
The assumption $P(1)=0$ implies $a+b+c+d=0$. There are four possibilities:
The first two cases can be dismissed because if all four coefficients are odd, then $P(2)$ is odd. The third case can be dismissed because no two powers of $2$ are equal, and the fourth case can be dismissed because no two powers of $2$ sum to another power of $2$.
Note, this proof applies, with a tiny extra argument, to any polynomial with at most $4$ nonzero (integer) coefficients. (The extra argument is to factor out and ignore the largest power of $x$ dividing the polynomial.)