1. Given any positive integer $a$, define the integer sequence $x_{1}, x_{2}$, $\cdots$, satisfying
$$
x_{1}=a, x_{n}=2 x_{n-1}+1(n \geqslant 1) .
$$

If $y_{n}=2^{x_{n}}-1$, determine the maximum integer $k$ such that there exists a positive integer $a$ for which $y_{1}, y_{2}, \cdots, y_{k}$ are all prime numbers.
1. If $y_{i}$ is a prime number, then $x_{i}$ is also a prime number.

Otherwise, if $x_{i}=1$, then $y_{i}=1$ is not a prime number; if $x_{i}=m n($ integers $m, n>1)$, then
$\left(2^{m}-1\right) \mid\left(2^{x_{i}}-1\right)$, i.e., $x_{i}$ and $y_{i}$ are composite numbers.

Below, we prove by contradiction: For any odd prime $a$, at least one of $y_{1}, y_{2}, y_{3}$ is composite.
Otherwise, $x_{1}, x_{2}, x_{3}$ are all primes.
Since $x_{1} \geqslant 3$ is odd, we know
$x_{2}>3$, and $x_{2}=3(\bmod 4)$.
Therefore, $x_{3} \equiv 7(\bmod 8)$.
Thus, 2 is a quadratic residue of $x_{3}$, i.e., there exists $x \in \mathbf{N}_{+}$ such that
$$
x^{2}=2\left(\bmod x_{3}\right) \text {. }
$$

Therefore, $2^{x_{2}}=2^{\frac{x_{3}-1}{2}} \equiv x^{x_{3}-1} \equiv 1\left(\bmod x_{3}^{3}\right)$.
Thus, $x_{3} \mid y_{2}$.
Since $x_{2}>3$, then $2^{x_{2}}-1>2 x_{2}+1=x_{3}$.
Therefore, $y_{2}$ is composite.
Finally, if $a=2$, then
$$
y_{1}=3, y_{2}=31 \text {, and } 23 \mid y_{3}=\left(2^{11}-1\right) \text {. }
$$

Therefore, $k=2$.