Because $+\infty$ is an element of the extended real numbers. That number system is extremely useful in calculus and analysis, which is why people explain things using them.
Continuity is a key idea in these fields, and we often prefer to continuously extend functions whenever possible. e.g. we often like to continuously extend the function $f(x) = x/x$ to have the value $f(0) = 1$. And when we use the extended real numbers, we like to continuously extend all sorts of functions to have values at $+\infty$ and $-\infty$ when appropriate.
For any $x > 1$, you have $x^{+\infty} = +\infty$. This is right because, for example,
$$ x^{+\infty} = \lim_{\substack{y \to x \\ n \to +\infty}} y^n $$
And for $0 < x < 1$, you have $x^{+\infty} = 0$ for the same reason. (this post will not comment on defining $0^x$)
But because $1$ lies on the boundary between where the value of $x^{+\infty}$ should be $0$ and where it should be $+\infty$, you can't continuously extend the function $x^n$ to $x = 1$ and $n = +\infty$. Thus, we leave it undefined.
As an aside, if you were thinking specifically of $f(x) = 1^x$ rather than $g(x,y) = x^y$, then it would make sense to define $f(+\infty) = 1$, even though we need to leave $g(1, +\infty)$ undefined.