Does $\mathbb{E}\left[X\right]=\infty$ imply $\mathbb{E}\left[X^{2}\right]=\infty$?

By Jensen's inequality we know that for convex $f$

$$ f\left(\operatorname{E}[X]\right) \leq \operatorname{E}\left[f(X)\right] $$

The fact that $\operatorname{E}[X]=\infty \implies \operatorname{E}\left[ X^2 \right] = \infty$ follows from the observation that $f(x)=x^2$ is convex.

Edit: As referenced in the comments below, for this argument to be rigorous, one needs to multiply inside the expectation by $\mathbf 1(\vert X \vert \leq k)$, apply Jensen's inequality, and then take $k \to \infty$ with an appeal to the monotone convergence theorem.

Proof that if $E[X]=\infty$ then $E[X^2]=\infty$ :
First note that $\int_{-1}^1xf_X(x)dx<=1$ and so if $E[X]=\infty$ then $\int_{\infty}^{-1}xf_X(x)dx+\int_1^{\infty}xf_X(x)dx=\infty$ and so $$E[X^2]>=\int_{\infty}^{-1}x^2f_X(x)dx+\int_1^{\infty}x^2f_X(x)dx>=\int_{\infty}^{-1}xf_X(x)dx+\int_1^{\infty}xf_X(x)dx=\infty$$

Note that $$ 0\leq\text{Var} X=E(X-EX)^{2}=EX^2-(EX)^2 $$ so that $$ (EX)^2\leq EX^2. $$ Hence $$EX=\infty \implies EX^2=\infty.$$