$\renewcommand{\P}{\operatorname{\mathsf P}}
\newcommand{\al}{\alpha}
\newcommand{\be}{\beta}
\newcommand{\la}{\lambda}
\newcommand{\D}{\overset{\text{D}}=}
\newcommand{\eD}{\overset{\text{D}}\to}
\newcommand{\si}{\sigma}
\newcommand{\tZ}{\tilde Z}$

Let $(X_1,\dots,X_d)$ be the observed normal random vector. Then the Neyman--Pearson test will reject $H_0$ if $S:=\sum_1^d(X_i-k\mu_i)^2<c$, where $\mu=:(\mu_1,\dots,\mu_d)$ and $c$ is a critical value and
\begin{equation*}
k:=\frac{\tau^2}{\tau^2-\sigma^2}. \tag{0.1}
\end{equation*}
With $\al\in(0,1)$ fixed, for large $d$ the hypotheses $H_0$ and $H_1$ will be too easy to distinguish from each other (so that the power be close to $1$) unless $H_0$ and $H_1$ are close enough to each other. The following rather natural conditions will provide for that:
\begin{equation*}
d\to\infty,\quad\tau^2\to\tau_0^2\in(0,\infty),\quad \frac{\tau^2-\si^2}{\si^2}=\frac{\tau^2}{\si^2}-1\sim\frac{w^2}{\sqrt d},\quad \frac{\|\mu\|^2}{\si^2}\sim b^2 \tag{0.2}
\end{equation*}
for some fixed $w,b$ in $(0,\infty)$, so that
\begin{equation*}
\tau\sim\si,\quad k\sim\frac{\sqrt d}{w^2}\to\infty. \tag{0.3}
\end{equation*}

Let $Z,Z_1,Z_2,\dots$ be iid standard normal random variables.
Under $H_0$, by the spherical symmetry,
\begin{equation*}
S\D\sum_1^d(\tau Z_i-k\mu_i)^2
\D\tau^2((a_0+Z_1)^2+Z_2^2+\dots+Z_d^2),
% \eD\tau^2(a_0^2+d+Z\sqrt{2d}),
\end{equation*}
where $\D$ means the equality in distribution and
\begin{equation*}
a_0^2:=(k/\tau)^2\|\mu\|^2\sim k^2b^2\sim\frac{b^2}{w^2}\,d\to\infty. \tag{1}
\end{equation*}
By the central limit theorem,
\begin{equation*}
\frac{Z_2^2+\dots+Z_d^2-d}{\sqrt d}\eD Z\sqrt2,
\end{equation*}
where $\eD$ means the convergence in distribution. Also, by (1),
\begin{equation*}
\frac{(a_0+Z_1)^2-a_0^2}{\sqrt d}=\frac{2a_0Z_1+Z_1^2}{\sqrt d}\to\frac{2b}w\,Z_1
\end{equation*}
pointwise and hence in distribution. Thus, under $H_0$,
\begin{equation*}
(H_0):\quad \frac{S/\tau^2-(a_0^2+d)}{\sqrt d}\eD \frac{2b}w\,Z_1+Z\sqrt2\D\la Z,\quad \la:=\sqrt{\frac{4b^2}{w^2}+2}. \tag{2}
\end{equation*}

Similarly, under $H_1$,
\begin{equation*}
S\D\sum_1^d(\si Z_i-(k-1)\mu_i)^2
\D\si^2((a_1+Z_1)^2+Z_2^2+\dots+Z_d^2),
% \eD\si^2(a_1^2+d+Z\sqrt{2d}),
\end{equation*}
where
\begin{equation*}
a_1^2:=((k-1)/\si)^2\|\mu\|^2\sim a_0^2\sim\frac{b^2}{w^2}\,d\to\infty, \tag{3}
\end{equation*}
and hence, under $H_1$,
\begin{equation*}
(H_1):\quad \tZ:=\frac{S/\si^2-(a_1^2+d)}{\la \sqrt d}\eD Z; \tag{3.1}
\end{equation*}
cf. (2).

Choosing now the critical value
\begin{equation*}
c:=\tau^2(a_0^2+d+z_\al\la\sqrt{d}), \tag{3.2}
\end{equation*}
where $z_\al:=\Phi^{-1}(\al)$ and $\Phi$ is the standard normal cdf, we see that the size of the test is
\begin{equation*}
\P_{H_0}(S<c)=\P_{H_0}\Big(\frac{S-\tau^2(a_0^2+d)}{\la\sqrt d}<z_\al\Big)\to\P(Z<z_\al)=\al,
\end{equation*}
as desired.

Recalling (3.1), (3.2), (1), (3), (0.3), (0.2), the relation $k^2-(k-1)^2\sim2k$, and the definition of $\la$ in (2), for the power of the test we now have
\begin{align*}
\P_{H_1}(S<c)&=\P_{H_1}\Big(\tZ<z_\al\frac{\tau^2}{\si^2}
+\frac{\tau^2-\si^2}{\la\si^2}\,\sqrt d
+\frac{[k^2-(k-1)^2]\|\mu\|^2}{\la\si^2\sqrt{d}}\Big) \\
& \to\Phi\Big(z_\al+(w^2+2b^2/w^2)\Big/\sqrt{\frac{4b^2}{w^2}+2}\Big),
\end{align*}
which latter is $>\al$, and it is close to $1$ if either $w$ or $b$ is large, which seems to agree with the intuition.

For this choice of H0 and H1, computing the rejection region in closed form does not appear to be possible,Why? It is just a quadratic inequality with $|X|^2$ and $\langle X,\mu\rangle$. The more difficult question is, of course, how $c$ depends on $\alpha$, but if the dimension is really high, you can be pretty sure that the sections of the corresponding body of revolution are either of nearly $0$ measure, or of nearly full measure with rather sharp transition from one case to another, so you can approximate pretty well by a simple cutoff function. $\endgroup$ – fedja Nov 18 '18 at 21:52