Four. (50 points) Find all positive integers $n$, such that there exists an integer $m$, making $2^{n}-1$ a divisor of $m^{2}+289$.

保留源文本的换行和格式，翻译结果如下：

Four. (50 points) Find all positive integers $n$, such that there exists an integer $m$, making $2^{n}-1$ a divisor of $m^{2}+289$.

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Note: The translation preserves the original format and line breaks.
For $n=2^{k}$ (where $k$ is a non-negative integer).

On one hand, if $n$ has an odd factor $s>1$, then $2^{s}-1$ is a factor of $2^{n}-1=\left(2^{s}\right)^{\frac{n}{1}}-1$, i.e.,
$2^{n}-1=\left(2^{2}\right)^{\frac{n}{4}}-1$
$=\left(2^{s}-1\right)\left[\left(2^{s}\right)^{\frac{n}{4}-1}+\left(2^{s}\right)^{\frac{n}{3}-2}+\cdots+\left(2^{s}\right)+1\right]$.
If $2^{n}-1$ is a factor of $m^{2}+289$, then $2^{3}-1$ is a factor of $m^{2}+289$.
Since $3 \times\left(2^{x}-1\right), 17 \times\left(2^{5}-1\right)$, and
$$
2^{s}-1 \equiv-1(\bmod 4),
$$

Therefore, $2^{s}-1$ must have a prime factor $p>3$, and
$p \equiv -1(\bmod 4)$.
Thus, $m^{2} \equiv -289(\bmod p)$.
Raising both sides to the $\frac{p-1}{2}$ power gives
$$
m^{p-1} \equiv -17^{p-1}(\bmod p).
$$

By Fermat's Little Theorem, $1 \equiv -1(\bmod p)$, i.e., $0 \equiv 2(\bmod p)$, which is a contradiction.
On the other hand,
$$
\begin{array}{l}
2^{2 k}-1=\left(2^{2 k-1}+1\right)\left(2^{2 k-1}-1\right)=\cdots \\
=\left(2^{2 k-1}+1\right)\left(2^{2 k-2}+1\right) \cdots\left(2^{21}+1\right)\left(2^{20}+1\right).
\end{array}
$$

The congruence equation $x^{2} \equiv -1\left(\bmod \left(2^{j}+1\right)\right)$ has a solution
$$
x=2^{2^{j-1}}\left(\bmod \left(2^{2^{j}}+1\right)\right).
$$

When $j>i$,
$$
\begin{array}{l}
\left(2^{2 i}+1,2^{2 j}+1\right)=\left(2^{2^{i+1}}-1,2^{2 j}+1\right)=\cdots \\
=\left(2^{2^{j}}-1,2^{2 j}+1\right)=1.
\end{array}
$$

Therefore, according to the Chinese Remainder Theorem, the system of congruence equations
$$
x \equiv 2^{2^{j-1}}\left(\bmod \left(2^{2^{j}}+1\right)\right)(j=1,2, \cdots, k-1)
$$

has a solution $x_{0}$.
Let $m=17 x_{0}$. Then $m^{2}+289=289\left(x_{0}^{2}+1\right)$ is divisible by $2^{2^{k}}-1$.