= Solution
A positively oriented planar <rotation matrix> acts by
$$
\boxed{\begin{pmatrix}x_1'\\x_2'\end{pmatrix}
=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}
\begin{pmatrix}x_1\\x_2\end{pmatrix}.}
$$
For a <unit vector> $n$ in three dimensions, the parallel and perpendicular <orthogonal projections> are
$$
x_\parallel=(x\cdot n)n,\qquad x_\perp=x-(x\cdot n)n.
$$
On the plane perpendicular to $n$, the map $v\mapsto n\times v$ preserves lengths and turns vectors through a positive right angle about $n$. It squares to minus the identity there. Thus $\cos\theta\,x_\perp+\sin\theta(n\times x_\perp)$ rotates that component, while the parallel component remains fixed. This proves the <Rodrigues rotation formula>
$$
x'=(x\cdot n)n+\cos\theta\,[x-(x\cdot n)n]+\sin\theta(n\times x).
$$
Its component form uses the <Levi-Civita symbol>:
$$
\boxed{R_{ij}=\cos\theta\,\delta_{ij}+(1-\cos\theta)n_in_j-\sin\theta\,\epsilon_{ijk}n_k.}
$$
Since $n_in_i=1$ and $\epsilon_{iik}=0$, its <matrix trace> is $3\cos\theta+1-\cos\theta=1+2\cos\theta$.
For the given matrix, the <matrix trace> is $1+2r$ with $r=1/\sqrt2$, so $\cos\theta=1/\sqrt2$. In the stated angular range, $\theta=\pm\pi/4$. The antisymmetric entries give
$$
2\sin\theta\,n=(R_{32}-R_{23},R_{13}-R_{31},R_{21}-R_{12})=(1,1,0).
$$
Therefore the two axis-angle pairs are
$$
\boxed{\left(\frac\pi4,\frac{(1,1,0)}{\sqrt2}\right),\qquad
\left(-\frac\pi4,-\frac{(1,1,0)}{\sqrt2}\right).}
$$
Reversing the axis and the angle leaves the <Rodrigues rotation formula> unchanged, so these describe the same rotation.
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