### Claim.
For every positive integer $n \ge 3$ there exists a positive integer $k$ such that $51^{2^{n-2}} - 1 = 2^n \cdot (2k + 1)$.

*Proof.* We prove the claim by mathematical induction on $n$.

For $n=3$ we have $51^2 - 1 = 3^2 - 1 = 8$ (mod 16), so $51^2 - 1$ is divisible by $2^3$ but not by $2^4$, therefore the claim holds for $n=3$. Assume the claim holds for some $n$, i.e. $51^{2^{n-2}} - 1 = 2^n \cdot (2k + 1)$ for some positive integer $k$. Then
$$\begin{aligned} 51^{2^{n-1}} - 1 &= (51^{2^{n-2}} - 1)(51^{2^{n-2}} + 1) \\ &= 2^n \cdot (2k+1)(2^n \cdot (2k+1) + 2) \\ &= 2^{n+1} \cdot (2k+1)(2^{n-1}(2k+1) + 1), \end{aligned}$$
and since $(2k+1)(2^{n-1}(2k+1)+1)$ is odd, the claim holds for $n+1$. This completes the proof. $\square$

Regarding the initial statement of the problem, we will inductively define an appropriate $k_n$ for every $n$. Note that for $n=1, n=2$ and $n=3$ we can select $k=2$, i.e. $k_1 = k_2 = k_3 = 2$. Assume the claim holds for some $n$, i.e. there exists some positive integer $a_n$ such that $51^{k_n} - 17 = 2^n \cdot a_n$. If $a_n$ is even, we can select $k_{n+1} = k_n$. If $a_n$ is odd, we can define $k_{n+1} = k_n + 2^{n-2}$. Then we have
$$51^{k_{n+1}} - 17 = 51^{2^{n-2}} \cdot (51^{k_n} - 17) + 17 \cdot (51^{2^{n-2}} - 1).$$
By the inductive assumption and the previously proven claim, we can conclude that there exists a positive integer $k$ such that
$$\begin{aligned} 51^{k+n+1} - 17 &= 51^{2^{n-2}} \cdot 2^n \cdot a_n + 17 \cdot 2^n \cdot (2k+1) \\ &= 2^n \cdot (51^{2^{n-2}} a_n + 17(2k+1)). \end{aligned}$$
Since $a_n$ is odd, it follows that $51^{2^{n-2}} a_n + 17(2k+1)$ is even, therefore $51^{k+n+1} - 17$ is indeed divisible by $2^{n+1}$.