Is this a bijection? Let X be a set. All unions are disjoint unions.
Suppose $f:X \rightarrow \{ \emptyset \} \cup X \cup X^2$ is a bijection.
Is there a bijection $g: X \cup X^3 \rightarrow X^2 \cup X^4$?
My first idea is to notice $g$ is equal to $h: X \times \{ \emptyset \} \cup X^2 \rightarrow X^2 \times \{ \emptyset \} \cup X^2$.
So I can send the second coordinate to itself. But then I get stuck. The first function restricted to some subset of $X$ is a bijection with $X^2$ but I don't know where to send the rest of the points. Any ideas?
 A: The inverse of $f$ is a bijection $X^0\cup X\cup X^2\to X$, and restricting this to $X^2$, we get an injection $X^2\to X$.  The diagonal map $x\mapsto (x,x)$ is an injection $X\to X^2$, so by Schroder-Bernstein, there exists a bijection $X\to X^2$.  As you note, this can then be combined with the identity map $X^0\cup X^2\to X^0\cup X^2$ to give a bijection $X\cup X^3\to X^2\cup X^4$.
(Note that this argument does not require the axiom of choice, whereas simply deducing that $|X|=|X|^2$ from the fact that $X$ is infinite as in BrianO's answer does.)
A: Because $f$ is a bijection $X \rightarrow \{ \emptyset \} \cup X \cup X^2$, there is an injection from $X$ into a proper subset of itself $A = f^{-1}(X)$,. (The injection is $f|A$.)
Therefore $X$ is infinite. It follows that $|X| = |X^2| = |X^3| = |X^4|$ and so on. Thus there is a bijection, hence an injection, $g: X \cup X^3 \rightarrow X^2 \cup X^4$.
Of course, if you really only want an injection, that's easy, for any $X$:
$$
d\colon x\mapsto (x,x)\colon X\to X^2
$$
is an injection, and
$$
e\colon (x,y,z)\mapsto (x,y,z,x)\colon X^3\to X^4
$$
is also an injection, 
so 
$$
d\cup e\colon X\cup X^3 \to X^2 \cup X^4
$$
is an injection too.

Here's how to express the bijection in terms of $f$.
First, $f$ gives a bijection
$$\begin{align}
h_1\colon (w,x)\mapsto (w,f(x))\colon X^3 &\stackrel{\sim}\to X\times (\{ \emptyset \} \cup X \cup X^2)\\
&= X\times\{ \emptyset \} \cup X^2 \cup X^3 \\
&\cong X^1 \cup X^2 \cup X^3, \\
\end{align}$$
and thus a bijection
$$
h = h^{-1}_1\colon X^1 \cup X^2 \cup X^3 \stackrel{\sim}\to X^2.
$$

Next, $f$ gives a bijection
$$\begin{align}
h_2\colon (w,x,y)\mapsto (w,f(x),y)\colon X^3 &\stackrel{\sim}\to X\times (\{ \emptyset \} \cup X \cup X^2) \times X \\
&= X\times\{ \emptyset \}\times X \cup X^3 \cup X^4 \\
&\cong X^2 \cup X^3 \cup X^4, \\
\end{align}$$
and thus a bijection
$$
id_X \cup h_2\colon X\cup X^3 \stackrel{\sim}\to X\cup X^2\cup X^3\cup X^4.
$$
Now, applying $h$ to $X\cup X^2\cup X^3$ gives a bijection
$$
g\colon X\cup X^3 \stackrel{\sim}\to X^2 \cup X^4.
$$
