Bounding a $\Vert X\Vert_{op}^2$ term with probability $1$ with $X$ being i.i.d. and $\mathbb P(X_{11} =1)=\mathbb P(X_{11} =-1)=\frac{1}{2}$.

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Let $X$ be a $N \times N$ random matrix with $i.i.d.$ random entries, and $$\mathbb P(X_{11} =1)=\mathbb P(X_{11} =-1)=\frac{1}{2}$$ Define $$\Vert X\Vert_{op}=\sup_{\mathbf{v}\in\mathbb C^n:\Vert \mathbf{v}\Vert_2=1} \Vert X\mathbf{v} \Vert_2,$$ show that for any fixed $\delta > 0$, $$\lim_{N\rightarrow\infty} \mathbb{P}(\Vert X\Vert_{op}\ge N^{\frac{1}{2}+\delta}) = 0$$ $$$$ The $\Vert X\Vert_{op}$ term is new here, so a proper way of dealing it should be bounding $\Vert X\Vert_{op}$ with some other known terms. We know that we can use the $tr(X^HX)$ to bound the $\Vert X\Vert_{op}^2$, but the $tr(X^HX)=N^2$ and it didn’t use properties of $X$. Are there any other terms which can not only be used to bound $\Vert X\Vert_{op}^2$, but also can use properties of $X$ to derive the bound? Thanks!

Edit1. To use the property of $X$, we hope to construct terms like “$X^2$”, not “$X^HX$”. Apart from $tr(X^HX)$, I hope that in this problem we can have something like $\Vert X\Vert_{op}^2 \le tr(X^2)$, but this is not true, are there any other ways of constructing bounds involving $X^2$?