The approximation
$$\sqrt{2} \approx \frac{3}{5} + \frac{\pi}{7 -\pi}$$
may be rewritten as
$$\sqrt{2}-\frac{3}{5} \approx \frac{1}{\frac{7}{\pi} -1}$$
After some manipulation, this is found to be equivalent to
$$\pi\approx\frac{392-175\sqrt{2}}{46}=\frac{7}{46}\left(56-25\sqrt{2}\right)$$
so, at least for this case, some theory relating $\pi$ to algebraic integers would suffice.
A useful direction is shown by the following series, which is related to a similar approximation to $\pi$:
$$\sum_{k=0}^{\infty} \frac{15!(k+1)}{(8k+1)_{15}}=\frac{15}{8}\left(1716-7\left(99\sqrt{2}-62\right)\pi\right)\approx 1$$
where $(a)_n$ is a rising factorial or Pochhammer symbol $a\times(a+1)\times...\times (a+n-1)$.
(see https://math.stackexchange.com/a/1657416/134791)
This gives the approximation
$$\pi \approx \frac{3676}{15(99\sqrt{2}-62)}=\frac{1838(62+99\sqrt{2})}{118185}$$
with eight correct decimal digits.
A general series that might provide an explanation for this approximation, as well as others of the form $a+b\sqrt{2}$ for rational $a$ and $b$, is given by
$$\sum_{k=0}^\infty \frac{c}{\prod_{i=1}^{7}((8k+i)(8k+16-i))^{w_i}} \approx 1,$$
with constant $c$ and binary weighting exponents $w_i$ taking values either $0$ or $1$. The example provided above is the particular case $w_i=1$ for all $i$ from $1$ to $7$.
The numerator $c$ may be set by letting the first term of the series equal $1$.
$$\sum_{k=0}^\infty \prod_{i=1}^7 \left(\frac{i(16-i)}{(8k+i)(8k+16-i)}\right)^{w_i} \approx 1,$$
A simple Dalzell-type integral that relates $\pi$ to approximations using $\sqrt{2}$ is given by
$$\pi=\frac{20\sqrt{2}}{9} - \frac{2\sqrt{2}}{3} \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2 + x^4 + x^6}dx$$
The following example combines a rational approximation from below and an irrational approximation from above
$$\pi \approx \frac{1}{4}·\frac{25}{8}+\frac{3}{4}·\frac{20\sqrt{2}}{9}$$ to obtain an irrational one from below that improves over the rational one.
$$\pi= \frac{25}{32} + \frac{5\sqrt{2}}{3} + \int_0^1 \frac{x \left(1 - x\right)^4\left(\left(1 + x^4\right) \left(1 + 4 x + x^2\right) - 8 \sqrt{2} x^3\right)}{16 \left(1 + x^2 + x^4 + x^6\right)} dx$$
WA link
Since the integrand is non-negative in $\left(0,1\right)$, this integral is a proof that
$$\pi > \frac{25}{32}+\frac{5\sqrt{2}}{3}$$
Hopefully a similar integral exists for your approximation. For instance, the integrand in
$$\frac{1}{184} \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2 + x^4 + x^6} (210\sqrt{2}- (259 + 120 x (1 - x)^2) (1+x^4) ) dx \\= \pi-\frac{392-175\sqrt{2}}{46}$$
is small, so your approximation is justified, but unfortunately there is a sign change in $(0,1)$, so we do not know in advance whether it lies above or below $\pi$. This integral has been built combining linearly three Dalzell-type expressions for constants $\frac{22}{7}-\pi$, $\frac{377}{120}-\pi$ and $\frac{20\sqrt{2}}{9}-\pi$ in order to match your number.
The relationship between the above families of series and integrals can be established as in the proofs found in these answers: [1],[2].
Finally, the following integral evaluates to the error of your approximation and has small nonnegative integrand.
$$\frac{3+ 10 \sqrt{2}}{4393} \int_0^1 \frac{x^3(1 - x)^6}{1 + x^2 + x^4 + x^6} \left(3111662 - 2200275\sqrt{2} + 3465 \left(-898 + 635 \sqrt{2}\right) x^8\right) dx = \pi-\frac{7}{46}\left(56-25\sqrt{2}\right)$$
It is a linear combination of
$$\frac{4}{10\sqrt{2}-3}\int_0^1 \frac{x^3(1-x)^6}{1+x^2+x^4+x^6}dx = \pi-\frac{35}{10\sqrt{2}-3}$$
and
$$\frac{4}{10\sqrt{2}-3}\int_0^1 \frac{x^{11}(1-x)^6}{1+x^2+x^4+x^6}dx = \pi-\frac{17327}{495\left(10\sqrt{2}-3\right)}$$
