Showing $64$ divides $5^n-8n^2+4n-1$ without induction

It seems to me that the simplest way is to use the binomial expansion $$ \begin{aligned} 5^n&=(4+1)^n=1+\binom n 1 4+\binom n2 4^2+\text{terms divisible by $4^3$}\\ &\equiv 1+4n+8n(n-1)\pmod{64}\\ &=1-4n+8n^2. \end{aligned} $$

Write $a_n=5^n-8n^2+4n-1 = 5^n+(-8n^2+4n-1)1^n$.

Then $a_n$ satisfies a linear recurrence implied by $(x-5)(x-1)^3$: $$ a_{n+4} = 8a_{n+3}- 18a_{n+2} + 16a_{n+1} - 5a_{n} $$

The particular expression for the recurrence is not important, except that it has integer coefficients.

Bottom line: It suffices to prove that $64$ divides $a_n$ for $n=0,1,2,3$. This is immediate because $a_0=a_1=a_2=0$ and $a_3=64$.