$\Theta(n^2 ) =\Theta(n^2 + 1)$
As we have on both sides of equality sets, then it should be prove correspondingly i.e we need to show, that $\Theta(n^2) \subset \Theta(n^2+1)$ and $ \Theta(n^2+1)\subset \Theta(n^2) $. Let's start with first:
Assume $f \in \Theta(n^2)$. This means, that for some constants $A_1, A_2 \gt 0$ holds
$$A_1 n^2 \leqslant f(n) \leqslant A_2 n^2$$.
To obtain inequalities for $\Theta(n^2+1)$ we need
$$B_1 (n^2+1) \leqslant f(n) \leqslant B_2 (n^2+1)$$
It's easy to see, that it's enough to take $B_2 = A_2$ for right hand and $0&lt;B_1 \leqslant \frac{A_1}{2}\leqslant \frac{A_1 n^2}{n^2+1}$ for left hand,  as $A_2&gt;0$.