Ring isomorphism $\phi:\Bbb Q[x]\to\Bbb Q[x]$ sending $\phi:x\mapsto (y+1)$ The question says:

"Show that the map $\phi:\Bbb Q[x]\to\Bbb Q[x]$ sending $\phi:x\mapsto (y+1)$ is a ring isomorphism."

$y$ is not defined anywhere. My question is, with superior knowledge of these sort of problems, what was meant to be asked? Alternatively what is being asked if it is written properly and I am just not getting it?

I assumed perhaps it was meant to be $x\mapsto x+1$
But as a ring homomorphism we need $\phi(x_1x_2)=\phi(x_1)\phi(x_2)$ but $(x_1+1)(x_2+1)\ne (x_1x_2+1)$
So that seems wrong.

I think it's a typo, it should be $x\mapsto x+1$, as you said. This is an homomorphism: 

if $p(x),q(x)\in \Bbb Q[x]$, then $\phi(p(x)q(x))$ is just the product
   $pq$ evaluated in $x+1$, which is the same as $p(x+1)q(x+1)$, i.e., $\phi(p(x))\phi(q(x))$. This is essentially the fact that the evaluation map is a ring homomorphism (whether you're evaluating elements of the ring or polynomials).

Considering Alex' comment, note you considered $x_1$ and $x_2$ as two different $x$'s, but there is only one indeterminant in $\Bbb Q[x]$, which is $x$, i.e., every element of this ring is a polynomial in one variable: $x$.
