Does the following hold for natural numbers:
$\forall x : \mathbb{N} , \forall y : \mathbb{N} \bullet (x < (y+1)^2) \Rightarrow ((x+1) < (y + 1 +1)^2)$ (1)
If (1) is valid I would like I to use it in a proof.
Is there some general theorem that states what transformations or functions can be applied to either side of an inequality?
Actually there doesn't exist such a theorem. it's because there are infinitely many transformations we can apply.
you can show easily in your case that the statement holds for every value of $x$ and $y$ in $\mathbb{N}$, namely:
Let $(1)$ be:
$$ x < (y+1)^2 $$
Let $(2)$ be:
$$ (x + 1) < (y + 2)^2 $$
If you apply $(2) - (1)$, you conclude that the statement holds for every $y$ greater than $-1$.
So the translation by $+1$ holds for every value in Natural Numbers. But there is no generalization. The most close would be the archimedean property of integers, if you had a linear expression.
But you need to precise the function between $x$ and $y$. not all translation by $1$ on $x$ gives you equivalent translation on $y$. here you lack of arguments.