There is a certain theorem that guarantees that the function value of the higher power of the Weber f function is an algebraic integer
\begin{align*} x\stackrel{?}{=}\mathfrak{f}^8(\tau)=\left(\frac{\eta\big(\frac{1+\tau}{2}\big)}{\exp\left(\frac{2 \pi\,\!i}{48}\right)\eta \left(\tau\right) }\right)^8 \end{align*}
where $\tau=\sqrt{-74}$.
I can't confirm whether the following polynomial is its minimum polynomial. Is there any algebraic verification method?
(x^10 - 8184*x^9 + 107568*x^8
- 4116480*x^7 + 24214784*x^6
- 190388224*x^5 + 387436544*x^4
- 1053818880*x^3 + 440598528*x^2
- 536346624*x + 1048576)
Ref:
https://homepages.warwick.ac.uk/~masfaw/classinv.pdf
https://www.math.u-bordeaux.fr/~ybilu/algant/documents/theses/Sotakova.pdf
(Comment has been modified to an answer.)
I. Weber functions
There are in fact three Weber modular functions,
\begin{align} \mathfrak{f}(\tau) &= \frac{\eta\big(\frac{1+\tau}{2}\big)}{\exp\left(\frac{2 \pi\,\!i}{48}\right)\,\eta\left(\tau\right)}\\[4pt] \mathfrak{f}_1(\tau) &= \frac{\eta\big(\frac{\tau}{2}\big)}{\eta\left(\tau\right)}\\[4pt] \mathfrak{f}_2(\tau) &= \frac{\sqrt2\,\eta\big(2\tau\big)}{\quad\eta\left(\tau\right)} \end{align}
There are three because the $24$th powers of these (with the proper sign) are the three roots of the simple cubic,
$$j(\tau) = \frac{(u-16)^3}y$$
where $j(\tau)$ is the j-function and the three roots are $u = \mathfrak{f}(\tau)^{24},\, -\mathfrak{f}_1(\tau)^{24},\, -\mathfrak{f}_2(\tau)^{24}.$ Or,
$$\sqrt[3]{j(\tau)} = \frac{v^3-16}v\quad$$
and the three roots are $v = \mathfrak{f}(\tau)^{8},\, -\mathfrak{f}_1(\tau)^{8},\, -\mathfrak{f}_2(\tau)^{8}.$ The first two can be quite close numerically. For example,
\begin{align} \mathfrak{f}(\sqrt{-74})^8 &= 8170.896840743\dots\\ \mathfrak{f}_1(\sqrt{-74})^8 &= 8170.896840503\dots \end{align}
Compare to the larger real root $x$ of the OP's decic,
$$\qquad\qquad x= 8170.896840503\dots$$
Therefore the decic root should be equated to the second Weber function as $\color{blue}{x = \mathfrak{f}_1(\sqrt{-74})^8},$ and which can be tested correct for hundreds of decimal digits.
II. The decic
Also, there are several clues that the decic is not some random equation. First, using the transformation, $x\to 4y$, it can be simplified to the palindromic,
$$1 - 2046y + 6723y^2 - 64320y^3 + 94589y^4 - 185926y^5 + 94589y^6 - 64320y^7 + 6723y^8 - 2046y^9 + y^{10} = 0$$
Let $y+\frac1y = z$, and this can be further simplified to,
$$-61378 + 74425z - 56136z^2 + 6718z^3 - 2046z^4 + z^5 = 0$$
which has square discriminant $d = 2^{46}\times7^6\times652783^2\times\color{red}{74^2}$.
Second, this quintic has transitive group of order $10$, hence is solvable in radicals. That is not surprising since it involves the Weber modular function $\mathfrak{f}_1(\sqrt{-74})$.