For just answering the yes/no question, the easiest way is to use the Swiss knife of bijections, the Cantor-Schröder-Bernstein theorem, which just requires us to construct separate injections in each direction $\mathbb R\to\mathbb N\times\mathbb R$ and $\mathbb N\times\mathbb R\to\mathbb R$ -- which is easy:
$$ f(x) = (1,x) $$
$$ g(n,x) = n\cdot \pi + \arctan(x) $$
Because there is an injection either way, Cantor-Schröder-Bernstein concludes that a bijection $\mathbb R\to\mathbb N\times\mathbb R$ must exist.
If you already know $|\mathbb R\times\mathbb R|=|\mathbb R|$, you can get by even quicker by restricting your known injection $\mathbb R\times\mathbb R\to \mathbb R$ to the smaller domain $\mathbb N\times\mathbb R\to\mathbb R$ instead of mucking around with arctangents.