The "sophisticated" answer is known as "Bezout's theorem". If you have a curve of degree $m$ and a curve of degree $n$, then they intersect in at most $mn$ places. (A "curve of degree $m$" is a curve with its equation given by a polynomial of degree $n$.) So a conic (degree 2) and a line (degree 1) intersect at most $2 \times 1 = 2$ times.
But you have to be careful about how you count the intersections. It turns out that a tangent counts as two intersections. Roughly speaking, this is because if you move the tangent line a little you can make it intersect the conic twice.
The problem here is that you have to prove Bezout's theorem first, which is not exactly the easiest thing in the world.
The more elementary answer is to write out the equations. The conic has equation
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$
and the line has equation
$$y = mx + b$$
assuming it's not a vertical line. So you can substitute $y = mx+b$ into the equation for the conic and get
$$Ax^2 + Bx(mx+b) + C(mx+b)^2 + Dx + E(mx+b) + F = 0.$$
Then you get scared that expanding this out will be a pain. But you don't have to! Just observe that $x$ never occurs with power greater than 2. So it's a quadratic in $x$, and has it has at most two solutions.
If the line is a vertical line, then it's $x = k$ for some constant $k$, and substituting this into the conic gives a quadratic in $y$.