Why is the eigenvalue equation $ Ax = \lambda x $ nonlinear?

It sounds like you're interested in why "equation problems" $Ax=b$ are called linear while "eigenvalue problems" $Ax=\lambda\cdot x$ are called nonlinear. There are likely many reasons to use this language, but here's one that invokes the notion of a linear combinations.

Suppose that $x_1$ and $x_2$ are two solutions to "equation problems" $Ax_1=b_1$ and $Ax_2=b_2$. Consider an arbitrary linear combination $x$ of $x_1$ and $x_2$, so $x=c_1\cdot x_1+c_2\cdot x_2$. If we define $b=c_1\cdot b_1+c_2\cdot b_2$, then $$ Ax = A(c_1\cdot x_1+c_2\cdot x_2) = c_1\cdot Ax_1+c_2\cdot Ax_2 = c_1\cdot b_1+c_2\cdot b_2 = b $$ This illustrates that "linear combinations of solutions to equation problems are solutions to equation problems."

Now, suppose that $x_1$ and $x_2$ are two solutions to "eigenvalue problems" $Ax_1=\lambda_1\cdot x_1$ and $Ax_2=\lambda_2\cdot x_2$. Again consider an arbitrary linear combination $x=c_1\cdot x_1+c_2\cdot x_2$. This linear combination $x$ solves an eigenvalue problem if $Ax=\lambda\cdot x$ for some $\lambda$. However, we have $$ Ax = A(c_1\cdot x_1+c_2\cdot x_2) = c_1\cdot Ax_1+c_2\cdot Ax_2 = c_1\cdot\lambda_1\cdot x_1+c_2\cdot\lambda_2\cdot x_2 \overset{?}{=} \lambda\cdot(c_1\cdot x_1+c_2\cdot x_2)=\lambda\cdot x $$ The $?$ in this equation indicates that a suitable $\lambda$ might not exist. Indeed, it isn't difficult to find exmamples where such a $\lambda$ does not exist. For instance, consider the data \begin{align*} A &= \left[\begin{array}{rr} 23 & 32 \\ -16 & -25 \end{array}\right] & \lambda_1 &= 7 & x_1 &= \left[\begin{array}{r} 2 \\ -1 \end{array}\right] & \lambda_2 &= -9 & x_2 &= \left[\begin{array}{r} 1 \\ -1 \end{array}\right] \end{align*} Note that $Ax_1=\lambda_1\cdot x_1$ and $Ax_2=\lambda_2\cdot x_2$. However, for $x=x_1+x_2$, we have $$ \overset{A}{\left[\begin{array}{rr} 23 & 32 \\ -16 & -25 \end{array}\right]}\overset{x}{\left[\begin{array}{r} 3 \\ -2 \end{array}\right]} = \left[\begin{array}{r} 5 \\ 2 \end{array}\right] \neq \lambda\cdot x $$

Side Note. As mentioned in the comments, there are other reasons to call the eigenvalue problem $Ax=\lambda\cdot x$ nonlinear. For example, if one uses the characteristic polynomial $\chi_A(t)=\det(t\cdot I_n-A)$ to solve for the eigenvalues of $A$, then one ends up factoring an $n$th degree polynomial, which is a nonlinear problem.

It's also worth noting that taking any linear combination of two eigenvectors $x_1$ and $x_2$ corresponding to the same eigenvalue $\lambda$ does yield a solution to an eigenvalue problem. This is precisely the statement that eigenvectors corresponding to an eigenvalue $\lambda$ are organized into the eigenspace $E_\lambda=\operatorname{Null}(\lambda\cdot I_n-A)$.

The most remarkable thing about the eigenvalue equation $$AX=\lambda X ~~~~(1)$$ is that the eigen vector $X$ very specially comes back on the RHS with a multiplicative constant (a scalar called eigenvalue $\lambda$). On the other hand in an ordinary linear equation $$AX=Y ~~~~(2)$$ $X$ does not come back, it is some other vector (say) $Y$ which is never proprtional to $X$.

For the differential operator $A=\frac{d}{dx}$, $f(x)= e^{ax}$ is the only possible eigenfunction such that $$\frac{d}{dx}f(x)=a f(x),$$ where $a$ is the eigenvalue. Note the special status of the the exponential function here. Other functions $\sin ax, \cos ax, \tan ax, e^{-ax^2}$ cannot be the eigenfunction of this $A$, because they do not come back by differentiating them.

The operator $A=\frac{d^2}{dx^2}$ being second order has two linearly independent (LI) degenerate eigenfunctions as $\sin ax$ and $\cos ax$ with the same eigenvalue $-a^2$. Importantly any linear combination of these rwo $$X= A \sin ax + B \cos ax$$ is also an eigenfunction. This is why (1) is linear whereas $$AX=\lambda X^2$$ is nonlinear.

Also it has two more LI degenerate eigenfunctions $e^{iax}$ and $e^{-iax}$ with the same eigenvalue are eigenfunctions of $\frac{d^2}{dx^2}$ and so is any linear combination of them $$X=C e^{-iax} + D e^{iax}$$ is also its eigenfunction.

In case of several pairs of degenerate eigenfunctions any member of one pair is a linear combination of the members of other pair, for instance $\cos ax= \frac{1}{2} e^{iax} +\frac{1}{2} e^{-iax}.$

Suppose

$A \in M_{n \times n}(\Bbb F), \; 0 \ne x \in \Bbb F^b, \; 1 \le n \in \Bbb N, \tag 0$

for some field $\Bbb F$.

With

$Ax = \lambda x, \; x \ne 0, \tag 1$

we have

$(A - \lambda I)x = Ax - \lambda x = 0, \tag 2$

whence

$x \in \ker (A - \lambda I) \Longrightarrow \det(A - \lambda I) = 0. \tag 3$

Now

$\det(A - xI) \in \Bbb F[x], \tag 4$

and by virtue of (0), we have

$\deg(A - xI) = n; \tag 5$

thus for $n > 1$, $\lambda$ satisfies a polynomial of degree greater than one, hence the equation (3) for $\lambda$ is non-linear.

Of course, the mapping

$A:\Bbb F^n \to \Bbb F^n, \; x \to Ax \tag 6$

is linear in $x$; however, the mapping

$\lambda x: \Bbb F \times \Bbb F^n \to \Bbb F^n, \; (\lambda, x) \to \lambda x \tag 7$

is non-linear when seen as a function of both $\lambda$ and $x$, hence neither is the mapping

$(A - \lambda I): \Bbb F \times \Bbb F^n \to \Bbb F^n, \; x \to (A - \lambda I)x; \tag 8$

the process of forming $\det(A - \lambda I)$ transforms these higher-dimensional non linearities into a (typically non-linear) polynomial, which is often more tractable in terms of locating those $\lambda \in \Bbb F$ for which (2) binds.