Let $X$ be a $n\times n$ matrix. I think that $\|X\|_2^2 = \|X^{\top}X\|_2$, because:
$$\|X^{\top}X\|_2^{2} = \max_{\|x\|_{2}=1} x^{\top}\left(X^{\top}X\right)^2x=\rho\left(\left(X^{\top}X\right)^2\right)=\left(\rho\left(X^{\top}X\right)\right)^2 = \|X\|_{2}^{4}$$
Where $\rho$ denotes the maximum eigenvalue / spectral radius of $X$.
Is this calculation correct?
You can use formulas $$\lVert A \rVert^2 = \sup_{\lVert u \rVert = \lVert v \rVert = 1}(Au, Av),$$ $$\lVert A \rVert = \sup_{\lVert u \rVert = \lVert v \rVert = 1}(Au, v),$$ both of which are consequences of Cauchy-Schwarz inequality. To get $\lVert X^TX \rVert = \lVert X \rVert^2$, use the above formulas and note that $(Xu, Xv) = (X^TXu, v)$.
The above argument works on an arbitrary Hilbert space.