What is wrong with this "proof"? $1$ is always an eigenvalue for $I + A$ ($A$ is nilpotent)? 
Consider the nilpotent matrix $A$ ($A^k = 0$ for some positive $k$).
  It is well known that the only eigenvalue of $A$ is $0$.
Then suppose $\lambda$ is any eigenvalue of $I + A$ such that 
  $(I + A) \mathbf{v} = \lambda \mathbf{v}$ where ($\mathbf{v} \neq \mathbf{0}$).
Then $I \mathbf{v} + A \mathbf{v} = \lambda \mathbf{v} \implies 
 \mathbf{v} + A \mathbf{v} = \lambda \mathbf{v} \implies A \mathbf{v} =
(\lambda - 1) \mathbf{v}$
We know that $\lambda - 1 = 0$ because the only eigenvalue of a
  nilpotent matrix is $0$. Therefore $\lambda = 1$

This "proof" seems to indicate that $1$ is always an eigenvalue for the sum of the identity matrix with any nilpotent matrix, but I believe I have a counterexample that disproves this.
I believe my error was in assuming that $I + A$ has eigenvalues -- but I do not know how I could prove/disprove this.
If someone could help me see where I've gone wrong I would greatly appreciate it!
 A: You have proven that if $A$ is nilpotent, then the eigenvalue of $A+I$ (NOT the eigenvalues of $A$) is equal to $1$. There's nothing wrong with the proof.
A: Here's an alternate argument (assuming the base field is algebraically closed):


*

*For a nilpotent matrix with $A^k=0$, all eigenvalues are $0$.
Proof:  Let $\lambda$ be an eigenvalue for $A$ and $v\not=0$ an eigenvector for $\lambda$.  Then
$$
0=0v=A^kv=\lambda^kv
$$
Since $v\not=0$, it must be that $\lambda^k=0$ so that $\lambda=0$

*$1$ is an eigenvalue for $I+A$.  Let $v$ be an eigenvector for $A$.  
Proof: We know that all eigenvalues for $A$ are $0$, so $Av=0$.  Therefore,
$$
(I+A)v=Iv+Av=v+0=v.
$$
Hence, $1$ is the eigenvalue for $v$ in $I+A$.

*You can prove more, every eigenvalue of $I+A$ is $1$.
Proof: Suppose that $\mu$ is an eigenvalue for $I+A$ and $w$ the corresponding eigenvector.  Then
$$
Aw=((I+A)-I)w=(I+A)w-Iw=\mu w-w=(\mu-1)w.
$$
Therefore, $w$ is an eigenvector for $A$ with eigenvalue $\mu-1$.  Since the only eigenvalues for $A$ are $0$, $\mu=1$.
A: If $\lambda$ is  an eigenvalue of a matrix $A$  then $p(\lambda)$ is eigenvalue of a polynomial $p(A)$.
In this case we have polynomial $p(A)=A+I$ so in this case the eigenvalues of $A$ (all zeros) should be shifted by $1$.
A: Your proof is right.

If $\lambda$ is an eigenvalue for the matrix $A$, then, for every scalar $\alpha$, $\lambda+\alpha$ is an eigenvalue for $A+\alpha I$.

Indeed, if $v\ne0$ and $Av=\lambda v$, then
$$
(A+\alpha I)v=Av+\alpha v=\lambda v+\alpha v=(\lambda+\alpha)v
$$
Now you can apply this to show that if $A$ has only the eigenvalue $0$, then $A+I$ has only the eigenvalue $1$. Conversely, if $A+I$ has only the eigenvalue $1$, then $A=(A+I)-I$ has only the eigenvalue $1-1=0$.
This doesn't depend on the field of scalars. However a real matrix is nilpotent if and only if its only eigenvalue over the complex numbers is $0$. As an example, the matrix
$$
\begin{bmatrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
$$
has $0$ as its only real eigenvalue, but is not nilpotent.
