Saying that something is at infinity is a convenient way of saying that the distance involved is much, much greater than the focal length.
So when the object is placed in the focal plane of the lens/mirror it is convenient to say that the image is formed at infinity.
In fact, in theory, the image is actually in two places.
On a screen as a real inverted image a long way away from the lens and on the opposite side of the lens to the object.
A virtual upright image a long way away from the lens which can be seen by looking through the lens at the object.
How can one infer all this?
By having the object close to the focal plane but remote from the lens and observing the image formed on a screen as being real, inverted and magnified.
Then moving the object closer and closer to the focal plane, seeing what happens and then imagining what would happen if the object was in the focal plane.
In the end all of this is theoretical because real lenses have defects and your analysis has dealt with ideal situations.
Any discussion of concave/convex mirrors needs to begin with a statement of the particular version of the mirror equation to be used, along with the convention for setting and interpreting the signs of focal lengths, and object/image positions.
For example, from http://scienceworld.wolfram.com/physics/MirrorFormula.html:
Unmentioned in this is the convention that virtual images and objects are found behind the mirror and have negative values of $d$.
In the particular example you present, the image formed by the primary mirror is a real image. If you put infinity for the object distance and a positive focal length, you find a positive image distance.
But when you insert a convex mirror, with a negative focal length, into the optical path, you must also consider the position of the real image (now an object) relative to the convex mirror. The object is behind the convex mirror; it is a virtual object, and its distance from the convex mirror is negative.
With appropriate positioning of the convex mirror, the formula will produce a positive value for the image distance. There will be real image formed in front of the convex mirror.
You've just designed a Cassegrain telescope...
Best Answer
A real image can be viewed on a screen, a virtual image can not. Rays from both types can enter your eye, be refracted by your eye's lens and form a real image on your retina as @CarlWitthoft points out.
So it is not the case that a real image must be viewed on a screen. It can be viewed on a screen.