This is Ben's solution to the first part:

Suppose $(x_1,y_1)$ is a point on the parabola $y={\mathrm{ax}}^2$ where $x_1$ and $y_1$ are integers, that is $y_1={\mathrm{ax}}_1^2$. Then, if $n$ is an integer, ${\mathrm{nx}}_1$ and $n^2$ are also integers and so

${\displaystyle a({\mathrm{nx}}_1{)}^2=n^2({\mathrm{ax}}_1^2)=n^2{y}_1.}$

So $({\mathrm{nx}}_1,n^2{y}_1)$ is another solution with integer coordinates. As $n$ can take an infinite number of integer values, if there is at least one lattice point solution, there are an infinite number.

This is Ruth's solution to the second part:

On the hyperbola $x^2-y^2=84=(x+y)(x-y)$, for $x$ and $y$ to be integers, $(x+y)$ and $(x-y)$ have to be the same parity because $(x+y)+(x-y)=2x$ and, for the total of two numbers to be even, they either have to be both odd or both even.

As $(x+y)(x-y)=84$ at least one bracket has to be even. As we require diophantine solutions, both brackets must be even. The only factorisations of 84 into two even numbers are:

${\displaystyle 84=2\times 42=42\times 2=-2\times -42=-42\times -2=6\times 14=14\times 6=-6\times -14=-14\times -6.}$

Each of these gives a distinct solution so the 8 solutions are $x=\pm 22,y=\pm 20$ (4 solutions) and $x=\pm 10,y=\pm 4$ ( 4 solutions).

There are two lattice points on the hyperbola in the first quadrant: (10,4) and (22,20). The lattice points (10, -4) and (22, -20) are the reflections of these points in the $x$-axis. Also $x\ge \sqrt{8}4$ or $x\le -\sqrt{8}4$ so there are two branches of the hyperbola. The other four lattice points lie on the other branch of the hyperbola and are the reflections of these four points in the $y$-axis.