Let $P(x)$ be the polynomial in question and let it takes values equal to powers of 2 at the points $x=j,j+1,\dots,j+9.$ Consider the finite difference of order 2 and step 1 $$\Delta^2P(x):=P(x)-2P(x+1)+P(x+2)$$Since $P$ is a monic polynomial of degree 2, we have $\Delta^2 P(x)=2.$ On the other hand, putting $x=j,j+1,\dots,j+9$ the longest possible sequence $f(j),f(j+1),\dots $ of values of $f$ all of which are power of 2 is $$ 8,4,2,2,4,8 $$and we can only satisfy $\Delta^2 f(x)=2$ for $x=j,j+1,j+2,j+3.$

The condition that the quadratic is monic is a bit of a red herring, since it is actually completely irrelevant (although one can use it as is shown by the solution in #2), as is the condition that the powers of $2$ have non-negative exponent: Indeed, if we have any quadratic polynomial $ax^2+bx+c$, then the third finite difference $f(x+3)-3f(x+2)+3f(x+1)-f(x)$ vanishes. Now, suppose that we have nine consecutive integral points where the values are powers of $2$ (not necessarily $\ge 1$). Then w.l.o.g. there are five consecutive of them where we are strictly increasing, say $n,n+1,n+2,n+3, n+4$. Now, $f(n+3)=3f(n+2)-3f(n+1)+f(n)<3f(n+2)+f(n)$. If $f(n+2) \le \frac{f(n+3)}{4}$, this is a contradiction since also $f(n) \le \frac{f(n+3)}{8}$. Hence we must have $f(n+3)=2f(n+2)$. But now $f(n+4)=3f(n+3)-3f(n+2)+f(n+1)=3f(n+2)+f(n+1) \le \frac{3f(n+4)}{4}+\frac{f(n+4)}{8}$ which is absurd.