By rational root theorem and some quick modular arithmetic, $-\tfrac{5}{2},\tfrac{5}{3},-\tfrac{5}{6}$ and $\tfrac{5}{7}$ are the only possible roots. Clearly the roots cannot be positive, so we must only realistically worry about $-\tfrac{5}{2}$ and $-\tfrac{5}{6}$. Now note that for $k>2$ we have $\tfrac{65^k}{25}$ and $\tfrac{65^k}{65}$ with at least one digit not $0$ or $1$ (namely the last digit, which is $5$), and we check that $4(-2.5)^3+2(-2.5)^2+2(-2.5)+5\neq 0$ to conclude.

Please check my solution. Let $\frac{-p}{q}$ is root. $P(x)=a_nx^n+\cdots+a_1x+a_0=(qx+p)(b_{n-1}x^{n-1}+...+b_2x^2+lx+m)$ where $q,p,b_i,l,m$ are integers. $65^k=P(10)=(10q+p)(100Q+10l+m)$ $65^k$ ends with $25$ ,so $pm=5,pl+qm=2,q|a_n \to q <10$ Case 1: $p=1$ Then $10q+1|65^k \to 101>10q+1=13^a$ - has not solutions. Case 2: $p=5$ $10q+5|65^k \to 21>2q+1=5^a13^b$ has solutions $q=2,6$ Case 2.1 $p=5,q=2$ $m=1,l=0$ $P(10)=25(100Q+1)=65^k \to k=2$ but $100Q+1=13^2$ has not solutions. Case 2.2 $p=5,q=6\to m=1,l=-4/5$ -contradiction. So there are not rational roots.

So first of all $a_0 = 5$ let rational roots $\frac{p}{q}$ so $q \mid a_n$ so transform the given polynomial in integer root polynomial by multiplying by $a_n^n$ so $ P(x) = a_n(a_nx)^n + a_{n-1}a_n(a_nx)^{n-1} + ..... + 5a_n^n$ let $a_nx = y$ so new polynomial having integer roots is $Q(y)= a_ny^n + a_{n-1}a_ny^{n-1} +....+ 5 a_n^n$ so let the integer roots are $q_i$ so $q_1.q_2....q_n= 5a_n^{n-1}$ notice $\sum (q_i) = -a_{n-1}$= a one digit number. All roots of $Q(x)$ is negative so if n≥10 then $a_{n-1} \geq 10 $ (denoting mod value as all roots are negative) so n≤10 and notice that one root is $-5$ so then n cannot be greater than 5 as $a_n \geq 1$ then $a_{n-1} \geq 11$ so n≤4 hence as $65^2$ is a 4 digit no so $n=4$ u can easily check thia case then we are done