Given $x, y \in \mathbf{N}_{+}$. Find the minimum value of $\sqrt{512^{x}-7^{2}-1}$.

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Let $z=512^{x}-7^{2 y-1}$.
Obviously, $z \geqslant 0$, and
$z \equiv 1(\bmod 7), z \equiv 1(\bmod 8)$.
(1) $x=2 x_{1}-1$ is an odd number.

Then $z \equiv(-1)^{x}-1 \equiv-2 \equiv 1(\bmod 3)$.
Thus, $z \equiv 1(\bmod 3 \times 7 \times 8)$.
Let $z=1$, i.e., $512^{2 x_{1}-1}-7^{2 y-1}=1$. Then $7^{2 y-1}=512^{2 x_{1}-1}-1$.
Hence $51117^{2 y-1} \Rightarrow 7317^{2 y-2}$, which is a contradiction.
Therefore, $z \geqslant 3 \times 7 \times 8+1=169$.
Let $z=169$, i.e., $512^{2 x_{1}-1}-7^{2 y-1}=169$.
Solving this, we get $(x, y)=(1,2)$.
Thus, the smallest non-negative value of $z$ is 169.
(2) $x=2 x_{1}$ is an even number.

Then $z \equiv(-1)^{x}-1 \equiv 0(\bmod 3)$.
Thus, $z \equiv 57(\bmod 3 \times 7 \times 8)$.
Let $z=57$, i.e., $512^{2 x_{1}}-7^{2 y-1}=57$.
Let $512^{x_{1}}=s, 7^{y-1}=t$. Then $s^{2}-7 t^{2}=57$.

The fundamental solution of the equation $s^{2}-7 t^{2}=1$ is $8+3 \sqrt{7}$.
Let $s_{1}+t_{1} \sqrt{7}$ be the fundamental solution of $s^{2}-7 t^{2}=57$. Then all solutions of the equation are
$$
S_{n+1}+t_{n+1} \sqrt{7}=\left(s_{n}+t_{n} \sqrt{7}\right)(8+3 \sqrt{7}),
$$

i.e.,
$$
s_{n+1}=8 s_{n}+21 t_{n}, t_{n+1}=3 s_{n}+8 t_{n}.
$$

Solving the first equation for $t_{n}=\frac{s_{n+1}-8 s_{n}}{21}$ and substituting into the second equation, we get
$$
\begin{array}{l}
s_{n+2}=16 s_{n+1}-s_{n} \\
\Rightarrow s_{n+2}=-s_{n}(\bmod 16).
\end{array}
$$

The fundamental solution $s_{1}+t_{1} \sqrt{7}$ should satisfy
$$
\begin{array}{l}
0 \leqslant s_{1} \leqslant \sqrt{\frac{1}{2}(8+1) \times 57}, \\
0 \leqslant t_{1} \leqslant \frac{3 \sqrt{27}}{\sqrt{2(8+1)}},
\end{array}
$$

i.e., $0 \leqslant s_{1} \leqslant 16,0 \leqslant t_{1} \leqslant 5$.
Trying each one, we find only
$$
\left(s_{1}, t_{1}\right)=(8,1),(13,4).
$$

When $\left(s_{1}, t_{1}\right)=(8,1)$, by conclusion (1) we get
$$
s_{2}=85 \equiv 5(\bmod 16);
$$

When $\left(s_{1}, t_{1}\right)=(13,4)$, by conclusion (1) we get $s_{2}=188 \equiv 12(\bmod 16)$.
Thus, $s_{n} \neq 0(\bmod 16)$, which is a contradiction.
Therefore, $512^{2 x_{1}}-7^{2 y-1}=57$ has no positive integer solutions.
Thus, $512^{2 x_{1}}-7^{2 y-1} \geqslant 168+57$.
In summary, the minimum value of $\sqrt{512^{x}-7^{2 y-1}}$ is 13.