Is there a vector field that is equal to its own curl?
I'd like to offer a high-brow answer to this question. While I realize that this solution is likely to be above the OP's level, I'm hoping it'll be of some interest to others.
Claim: Let $u(x,y)$, $v(x,y)$, $w(x)$ be any real-analytic functions (defined on a neighborhood of $0$). Then there exists a unique real-analytic vector field $F = (F^1, F^2, F^3)$ that satisfies $\text{curl}\,F = F$ and $$\begin{align} F^1(x,0,0) & = w(x) \\ F^2(x,y,0) & = u(x,y) \\ F^3(x,y,0) & = v(x,y). \end{align}$$
We'll show this by applying the Cauchy-Kovalevskaya Theorem twice. Here I should acknowledge that I first saw this techinque applied to $\text{curl}\,F = F$ in an exercise in Robert Bryant's "Nine Lectures on Exterior Differential Systems."
Cauchy-Kovalevskaya Theorem: If $\mathbf{H}$ and $\phi$ are real-analytic functions near the origin, then there is a neighborhood of the origin on which there exists a unique real-analytic solution $\mathbf{g}(\mathbf{x}, t)$ to $$\begin{align} \frac{\partial \mathbf{g}}{\partial t}(\mathbf{x},t) & = \mathbf{H}(\mathbf{x},t, \mathbf{g}(\mathbf{x},t), \frac{\partial \mathbf{g}}{\partial x^i}) \\ \mathbf{g}(\mathbf{x}, 0) & = \phi(\mathbf{x}) \end{align}$$ A PDE system of this form will be called a "Cauchy problem."
Setup
The equation $\text{curl}\,F = F$ can be viewed as an overdetermined system of PDEs -- in fact, a system of 4 first-order quasilinear equations for 3 unknown functions. That is, if we write $F = (F^1, F^2, F^3)$, then the condition $\text{curl}\,F = F$ becomes: $$\begin{align} \frac{\partial F^2}{\partial z} - \frac{\partial F^3}{\partial y} & = F^1 \tag{1} \\ \frac{\partial F^3}{\partial x} - \frac{\partial F^1}{\partial z} & = F^2 \tag{2} \\ \frac{\partial F^1}{\partial y} - \frac{\partial F^2}{\partial x} & = F^3. \tag{3} \end{align}$$ The fourth equation is a hidden "integrability condition" of sorts: namely, any solution $F$ to $\text{curl}\,F = F$ must satisfy $\text{div}\,F = 0$. This gives us a fourth (first-order quasilinear) equation $$\frac{\partial F^1}{\partial x} + \frac{\partial F^2}{\partial y} + \frac{\partial F^3}{\partial z} = 0. \tag{4}$$
The Cauchy Problems
We can write this system as a sequence of two Cauchy problems as follows. Let $u(x,y)$, $v(x,y)$, and $w(x)$ be arbitrary real-analytic functions.
We first consider the problem of finding $g(x,y)$ such that $$\begin{align} \frac{\partial g}{\partial y} & = \frac{\partial u}{\partial x} + v(x,y) \\ g(x,0) & = w(x). \end{align}$$ By Cauchy-Kovalevskaya$^\dagger$, there exists a unique real-analytic solution $g(x,y)$.
Second, letting $g(x,y)$ be a solution as above, we consider the problem of finding $F^1, F^2, F^3$ such that $$\begin{align} \frac{\partial F^1}{\partial z} & = -F^2 + \frac{\partial F^3}{\partial x} \\ \frac{\partial F^2}{\partial z} & = F^1 + \frac{\partial F^3}{\partial y} \\ \frac{\partial F^3}{\partial z} & = -\frac{\partial F^1}{\partial x} - \frac{\partial F^2}{\partial y} \\ F^1(x,y,0) & = g(x,y) \\ F^2(x,y,0) & = u(x,y) \\ F^3(x,y,0) & = v(x,y) \end{align}$$ By Cauchy-Kovalevskaya again, there exists a unique real-analytic solution $F = (F^1, F^2, F^3)$.
By construction, this solution $(F^1, F^2, F^3)$ satisfies equations (1), (2) and (4). One can check (exercise!$^*$) that this $F$ necessarily satisfies equation (3) as well. This completes the proof. $\lozenge$
End notes
$^\dagger$ Admittedly overkill here
$^*$ Hint for Exercise: Consider $E(x,y,z) := \frac{\partial F^1}{\partial y} - \frac{\partial F^2}{\partial x} - F^3$. Check that $E(x,y,0) = 0$ and $\frac{\partial E}{\partial z} = 0$, and then apply uniqueness in Cauchy-Kovalevskaya.
Note that this exercise shows why we needed the first Cauchy problem at all: that is, we couldn't just choose $F^1(x,y,0) = g(x,y)$ completely arbitrarily!
To give a final answer, I've found that, in the standard unit coordinates, the field $$F(x,y,z) = (-\cos z, \sin z, 0)$$ is equal to its curl. Again, brought to you by this paper.
P.S. My first attempt revealed that Wolfram Alpha is a wise-guy.
if $f =(\sin z, \cos z, 0)\ $ then $\nabla \times f = f$