Notice that $\sqrt{51}\approx 7+\frac{\sqrt{2}}{10}$

Well, I was thinking this: $$\frac{d\sqrt{x}}{dx}|_{x=50} = \frac{1}{2\sqrt{x}}|_{x=50} = \frac{1}{10\sqrt{2}}.$$ So you'd expect the difference between $\sqrt{51}$ and $7=\sqrt{49}$ to be something very close to $2×\frac{1}{10\sqrt{2}} =\frac{\sqrt{2}}{10}$.

To approach with continued fractions:

$$\begin{align}a_n=\sqrt{n^2+2}&=[n,n,2n,n,2n,\dots]\\\sqrt{n^2+1}&=[n,2n,2n,\dots]\\ b_n=n+\frac{1}{\sqrt{n^2+1}}&=[n,n,2n,2n,2n,\dots] \end{align}$$

So the third continued fraction for both $a_n$ and $b_n$ are $$c_3=\frac{2n^3+3n}{2n^2+1}.$$ The fourth fraction for $a_n$ is: $$c_4=\frac{2n^4+4n^2+1}{2n^3+2n}.$$ We know $c_3<b_n<a_n<c_4,$ so $$0<a_n-b_n<c_4-c_3=\frac{1}{(2n^3+2n)(2n^2+1)}<\frac{1}{4n^5}$$

Your case is $n=7,$ because $\frac{\sqrt2}{10}=\frac1{\sqrt{7^2+1}}.$

If you have solutions of $n^2-2m^2=-1,$ you get the approximation: $$\sqrt{n^2+2}\approx n+\frac{\sqrt{2}}{2m}$$ There are infinitely many integer solutions $(n,m)$ computed as $n+m\sqrt2=(1+\sqrt2)(3+\sqrt2)^k.$

For example, $(n.m)=(7,5)$ or $(n,m)=(41,29).$ The latter gives:

$$\sqrt{1683}-41-\frac{\sqrt2}{58}=0.000000001077321\approx\frac12\cdot\frac{1}{4\cdot 41^5}$$

That $1/2$ is probably because the fourth coefficients of $a_n$ and $b_n$ are $n$ and $2n,$ respectively.

A little more work shows that $$a_n-b_n\sim\frac1{8n^5}$$

Specifically, I think you get: $$0<\frac{1}{2n(n^2+1)(4n^2+3)}-\left(a_n-b_n\right)<\frac{1}{32n^7}$$