One, (20 points) Given $t=\sqrt{2}-1$. If positive integers $a$, $b$, and $m$ satisfy
$$
(a t+m)(b t+m)=17 m
$$

find the value of $a b$.
Given that $t=\sqrt{2}-1$, we have
$$
t^{2}=3-2 \sqrt{2} \text {. }
$$

From the problem, we know
$$
\begin{array}{l}
a b t^{2}+m(a+b) t+m^{2}=17 m \\
\Rightarrow a b(3-2 \sqrt{2})+m(a+b)(\sqrt{2}-1)+m^{2}=17 m \\
\Rightarrow \sqrt{2}[m(a+b)-2 a b]+ \\
\quad\left[3 a b-m(a+b)+m^{2}-17 m\right]=0 .
\end{array}
$$

Since $a, b, m$ are positive integers, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
m(a+b)-2 a b=0, \\
3 a b-m(a+b)+m^{2}-17 m=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a+b=2(17-m), \\
a b=17 m-m^{2} .
\end{array}\right.
\end{array}
$$

Therefore, $a, b$ are the two integer roots of the quadratic equation in $x$:
$$
x^{2}+2(m-17) x+17 m-m^{2}=0
$$

The discriminant of equation (1) is
$$
\begin{array}{l}
\Delta=4(m-17)^{2}-4\left(17 m-m^{2}\right) \\
=4(17-m)(17-2 m) \geqslant 0 .
\end{array}
$$

Since $a, b, m$ are positive integers, we have
$$
a+b=2(17-m)>0 \text {. }
$$

Thus, $0<m \leqslant \frac{17}{2}$.
Since the discriminant $\Delta$ is a perfect square, upon verification, only $m=8$ meets the requirement.
Substituting $m=8$ into the equation, we get
$$
a b=17 m-m^{2}=72 \text {. }
$$