Is $L = \{(x,y,z) | x+y=z\}$ a regular language? Suppose $x,y,z$ are coded as decimal or their binary representations in an appropriate DFA. Is $L$ regular?
My intuition tells me that the answer is no, because there are infinitely many combinations such that $x+y=z$ and a DFA must contain a finite amount of memory. Is this correct?

Your question is rather imprecise, but it makes it even more interesting...
First, why is your question imprecise? Mainly because you not specify the alphabet and the coding of the triples $(x,y,z)$. So let me modify your question as follows:

Is there an encoding of the triples of integers $(x, y, z)$ on a
  suitable alphabet $A$ such that the set $L$ is a regular language of $A^*$?

The answer is yes if you take an appropriate coding. The first idea is to represent the natural numbers in reverse binary representation. For instance $22 = 2 + 4 + 16 \to
01101$, $13 = 1+4+8 \to 1011$ and $35 = 1 + 2 + 32 \to 110001$. Next you represent the triples $(x,y,z)$ of natural integers as in this example (note the zeros added at the end of the first two integers in order to have the same length for the representations of $x$, $y$ and $z$)
\begin{matrix}
  22 &\to &011010\\
  13 &\to &101100\\
  35 &\to &110001
\end{matrix}
Finally you take the following representation on the alphabet $A = \{0,1\}^3$.
$$
(22, 13, 35) \to
\begin{pmatrix}
  0\\1\\1
\end{pmatrix}
\begin{pmatrix}
  1\\0\\1
\end{pmatrix}
\begin{pmatrix}
  1\\1\\0
\end{pmatrix}
\begin{pmatrix}
  0\\1\\0
\end{pmatrix}
\begin{pmatrix}
  1\\0\\0
\end{pmatrix}
\begin{pmatrix}
  0\\0\\1
\end{pmatrix}
$$
Now I claim that the language $L$ of the representations on $A^*$ of all triples $(x, y, z)$ such that $x + y = z$ is regular and accepted by the following DFA (in which the letters are written as horizontal triples rather than as vertical ones to fit the picture on the screen)

Can you see why?
