Prove that for an integer $x \ge 7$, it follows that $x\# > x^2+x$

As an alternative, let's suppose we've verified the inequality for all integers up to $24$ (a simple extension of the OP's argument taking things up to $13$, since $2\cdot3\cdot5\cdot7\cdot11=2310\gt24^2+24$), and let's invoke Bertrand's Postulate in the equivalent form that for any real number $u\gt1$ there is a prime satisfying $u\lt p\lt2u$.

This version of BP allows for the following conclusion: If $x\gt24$, then there exist prime numbers $p$, $q$, and $r$ such that

$$2\lt3\lt{x\over8}\lt p\lt{x\over4}\lt q\lt{x\over2}\lt r\lt x$$

It follows that

$$\#x\ge6pqr\gt6\cdot{x\over8}\cdot{x\over4}\cdot{x\over2}={3x^3\over32}$$

and it's easy to see that

$${3x^3\over32}\gt x^2+x$$

if $x\gt24$ (in fact, if $x\ge12$).