$A$ is symmetric, nonnegative, and irreducible. By a theorem of Perron and Frobenius, $k$ is a simple eigenvalue with a positive eigenvector $u$. Now with componentwise absolute value, $k|x|=|-kx|=|Ax|\le A|x|$. Multiplication with $u^T$ shows that we must have equality. Hence $|x|$ is an eigenvector, hence a multiple of $u$. Therefore $x$ has no zero component.
Partition the nodes into $P$ (of size $p$) and $N$ (of size $n$), where $x_i>0$ if $i\in P$ and $x_i<0$ if $i\in N$. Then $v=x_P>0$ and $w=-x_N>0$. Partition $A$ conformally as $A=\pmatrix{B & C\\C^T & D}$ (of size $p+n\times p+n$, with $B$ of size $p\times p$, and note that $B,C,D$ are nonnegative. Then the equation $|Ax|= A|x|$ implies $|Bv-Cw|=Bv+Cw$ and $|C^Tv-Dw|=C^Tv+Dw$. Taking the squared norm and simplifying yields $(Bv)_i(Cw)_i=0$ for $i\in P$ and $(C^Tv)_k(Dw)_k=0$ for $k\in N$. Since $v,w>0$, $C_{ik}=1$ implies that $B_{ij}=0$ for $j\in P$ and $D_{kj}=0$ for $j\in N$.
This means that for every edge $ik$ with $i\in P$ and $k\in N$, the neighbors of $i$ must lie in $N$ and the neighbors of $k$ must lie in $P$. Growing the graph starting with some such edge implies that its connected component is bipartite. On the other hand, if there is no such edge then $P$ and $N$ are unions of connected components. Since the graph was assumed connected, it follows that it is bipartite.