On
Cartesian product, $$ (\mathbb{Z}^{+}) \times (\mathbb{Z}^{+}) \times (\mathbb{Z}^{+}) \times (\mathbb{Z}^{+}), $$ normally abbreviated as $(\mathbb{Z}^{+})^4$.
On
The set of all $4$-tuples of positive integers.
A single element is a $1$-tuple, an ordered pair is a $2$-tuple, and an ordered foursome is a four-tuple. So it's the set:
$$\left({\mathbb Z^+}\right)^4 = \left \{{(a,b,c,d): a, b, c, d \in \mathbb Z^+}\right\}$$
Where $(a,b,c,d) = (e,f,g,h)$ if and only if $a = e, b = f, c = g, d = h$.
You can also think of it as the set of all sequences of length four whose image is positive integers. This would be a function that maps $0 \mapsto a ,1 \mapsto b, 2 \mapsto c$, and $3 \mapsto d$.
The notation $(\mathbb Z^+)^4$ is just the set of tuples of $4$ positive integers. So, it says that $(a,b,c,n)$ is a tuple of $4$ positive integers. It would be equally clear to write $a\in\mathbb Z^+,b\in\mathbb Z^+,c\in\mathbb Z^+,n\in\mathbb Z^+$ or even to write $a,b,c,n\in\mathbb Z^+$ - all of which are equivalent. The author is likely looking to emphasize that Fermat's theorem is a statement about the non-existence of such a $4$-tuple by writing it that way - i.e. they're making the domain very explicit.