Step 1: Locate the point $P$.
For $x^2-y^2=1$ a point is $(\sec\phi,\tan\phi)$. At $\phi=\frac{\pi}{4}$ we get $P = (\sec\frac{\pi}{4},\tan\frac{\pi}{4}) = (\sqrt2,\,1)$.
Step 2: Find the slope of the normal at $P$.
Differentiating $x^2-y^2=1$ gives $\frac{dy}{dx} = \frac{x}{y}$, so at $(\sqrt2,1)$ the tangent slope is $\sqrt2$. The normal slope is therefore $-\frac{1}{\sqrt2}$.
Step 3: Write the normal line.
The normal through $(\sqrt2,1)$ is $y - 1 = -\frac{1}{\sqrt2}(x - \sqrt2)$, which simplifies to $x + \sqrt2\,y = 2\sqrt2$.
Step 4: Use the point $Q(\theta)=(\sec\theta,\tan\theta)$.
Since $Q$ lies on the normal, $\sec\theta + \sqrt2\tan\theta = 2\sqrt2$. The point $Q$ already satisfies the curve. We need $\sec^2\theta + \tan\theta$.
Step 5: Solve for $\tan\theta$.
From $\sec\theta = 2\sqrt2 - \sqrt2\tan\theta$ and $\sec^2\theta - \tan^2\theta = 1$: $(2\sqrt2-\sqrt2\tan\theta)^2 - \tan^2\theta = 1$, i.e. $8 - 8\tan\theta + 2\tan^2\theta - \tan^2\theta = 1$, so $\tan^2\theta - 8\tan\theta + 7 = 0$. This gives $\tan\theta = 7$ (taking the new point, since $\tan\theta=1$ is $P$). Then $\sec^2\theta = 1 + \tan^2\theta = 50$.
Step 6: Combine the required quantity.
Therefore $\sec^2\theta + \tan\theta = 50 + 7 = 57$.
\[ \boxed{57} \]