When a star becomes a black hole, does its gravitational field become stronger?

Actually, it doesn't have the same mass, it has significantly less mass than its precursor star. Something like 90% of the star is blown off in the supernova event (Type II) that causes the black holes.

The Schwarzschild radius is the radius at which, if an object's mass where compressed to a sphere of that size, the escape velocity at the surface would be the speed of light $c$; this is given by $$ r_s=\frac{2Gm}{c^2} $$ For a 3-solar mass black hole, this amounts to about 10 km. If we measure the gravitational acceleration from this point, $$ g_{BH}=\frac{Gm_{BH}}{r_s^2}\simeq10^{13}\,{\rm m/s^2} $$ and compare this to the acceleration due to the precursor 20 solar mass star with radius of $r_\star=5R_\odot\simeq7\times10^8$ m, we have $$ g_{M_\star}=\frac{Gm_\star}{r_\star^2}\simeq10^3\,{\rm m/s^2} $$ Note that this is the acceleration due to gravity at the surface of the object, and not at some distance away. If we measure the gravitational acceleration of the smaller black hole at the distance of the original star's radius, you'll find it is a lot smaller (by a factor of about 7).

When you watch a pop-sci TV show, you need to take everything you see with a very healthy grain of salt. This is particularly the case if the show's host isn't a scientist, but even when a scientist is the host, you need to be suspicious.

Stellar black holes do not turn into monsters that reach out and pluck objects from the heavens. From far away, a black hole behaves no differently gravitationally than an ordinary object of equal mass. It's only when an object gets very close that black holes behave differently. Note that this "very close" means what would be well into the interior of the ordinary object.

If anything, stellar black holes are little kitty cats rather than big monsters compared to the stars that generated them. The supernovae that generate stellar black holes blow away a very large portion their mass during the supernova event, both as energy and ejected matter. The resulting black hole has a much smaller mass than did the parent star.

If the parent star is a member of a close binary star, the black hole might still be able to draw mass from the other star. But reaching out and inhaling planets? That's just bad pop-sci.

Except for the outer atmosphere of a red giant that is a close binary pair of a stellar black hole, it would be quite amazing for a stellar black hole to gobble up anything. It would take a lot of energy to intentionally send something very close to a black hole. By way of analogy, mankind has sent four satellites out of the solar system (with a fifth on the way) but we have only sent two missions to Mercury. The reason is that takes a lot of energy (a whole lot of energy!) to get to Mercury. Escaping the solar system is a piece of cake compared to getting to Mercury. It would take even more energy to get very close to the surface of the Sun. If our Sun was instead a one solar mass black hole, it would take a whole, whole lot more energy to send something within a few Schwarzschild radii of the black hole.

It actually goes the other way around: when a star collapses to form a black hole, its planets (if it has any) will become unbound and fly away to infinity.

Simple reason: when the star explodes to form a compact object (neutron star or black hole), it releases most of its mass in the form of a SuperNova explosion, so that the central object around which the planet is orbiting has a much smaller mass than the original star. The least decrease is roughly from an $8 M_\odot$ star to a $1.4 M_\odot$ neutron star, giving a minimum reduction of about a factor of 6.

Now let us consider what happens to the planet. Before the explosion, its kinetic energy $K$ is half the modulus of its potential energy $W$: $$ K = -\frac{1}{2} W $$ so that its total energy $E = T+W = -T/2 < 0$, and the planet is bound to the star.

But after the explosion, while the planet speed is left unchanged, its potential energy $W = -GM_\star M_{planet}/r$ is reduced because $M_\star$ has decreased by at least a factor of $6$: the new potential energy $-W_{final} < -W_{initial}/6$. Hence the new energy

$$ E = T + W_{final} = -\frac{1}{2}W_{initial} + W_{final} > -\frac{1}{3} W_{initial} > 0\;: $$

the final , total energy is positive, the planet is unbound from the star, it will just fly away from it.