1.) $Ax = \lambda x$, given.
2.) Inductive hypothesis: $A^kx = \lambda^k x$ for some positive integer $k$.
3.) Operate on the equation of (2) with $A$: $A^{k + 1}x = A(A^k x) = A(\lambda^k)x = \lambda^k (Ax) = \lambda^{k + 1} x$.
4.) Conclude that $A^k x = \lambda^k x$ for all positive integers $k$.
QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
The $2 \times 2$ case is quite simple: such an $A$ causes a shear transformation: one vector is fixed, the one perpendicular to it maps to one with the same perpendicular component, but the parallel component is changed (think of the effect on the basis consisting of $(1,0)$ and $(0,1)$: $(1,0) \mapsto (1,0)$ and so is unchanged, whereas $(0,1) \mapsto (1,1)$.
The general case of only having one eigenvector is similar: in the two-dimensional case, the possible Jordan Normal Forms are
$$ \begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix}, \quad \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}, \quad \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}, $$
where $\lambda \neq \mu$, and it is a significant theorem that any $2 \times 2$ matrix is related to one of these by a change-of-basis.
Now, the characteristic polynomial of any matrix of the first type is $(t-\lambda)(t-\mu)$, and its coefficients are real if the original matrix is real, so they are either both real or a complex conjugate pair. The complex case can be shown to be a rotation matrix, which has no real eigenvectors in two dimensions. The real case is a pair of unequal scalings.
The second case is a simple scaling, and it is easy to show that the only matrices of this type are of the form $\lambda I$ for real $\lambda$.
The last case is what your $A$ falls into, and it is easy to see that these are all shears, in the same way as $A$ is, but with additional scaling (since it factors into $\lambda I \begin{pmatrix} 1 & 1/\lambda \\ 0 & 1 \end{pmatrix}$).
Best Answer
Given $A=SBS^{-1}$ and
$$ (\lambda I-A)x=0 \implies (\lambda I- SBS^{-1} )x=0 \implies (\lambda S - SB )S^{-1}x=0 $$
$$ \implies S(\lambda I - B )S^{-1}x=0 \implies (\lambda I - B )S^{-1}x=S^{-1}0=0 .$$
The last equation implies that $B$ has the same eigenvalue as $A$ (which is a fact for the similarity matrices) with the eigenvector $ z=S^{-1}x .$