I know that $\sum a_{n}$, $\sum b_{n}$, $\sum (a_{n}^2 - b_{n}^2)$ and $\sum 2a_{n}b_{n}$ converges. With that can I say that $\sum a_{n}^{2}$ and $\sum b_{n}^{2}$ converges?
PS. $a_{n}$ and $b_{n}$ are from $\sum z_{n}$(converge) where $z_{n} = a_{n} +ib_{n}$ and the other series are from the hipothesis that $\sum z_{n}^2$ converges and $a_{n} \geq 0$
On
Utilizing the PS, let $z_n$ be the sequence
$$1,i,-1,-i,{1\over\sqrt2},{i\over\sqrt2},{-1\over\sqrt2},{-i\over\sqrt2},{1\over\sqrt3},{i\over\sqrt3},{-1\over\sqrt3},{-i\over\sqrt3},{1\over\sqrt4},{i\over\sqrt4}\ldots$$
so that $z_n^2$ is the sequence
$$1,-1,1,-1,{1\over2},{-1\over2},{1\over2},{-1\over2},{1\over3},{-1\over3}\ldots$$
Both $\sum z_n$ and $\sum z_n^2$ sum to $0$, but
$$\sum|z_n|^2=1+1+1+1+{1\over2}+{1\over2}+{1\over2}+{1\over2}+{1\over3}+{1\over3}+\cdots$$
diverges.
Teasing this apart, we have the convergent sums
$$\begin{align} \sum a_n&=1+0-1+0+{1\over\sqrt2}+0-{1\over\sqrt2}+0+{1\over\sqrt3}+0+\cdots\\ \sum b_n&=0+1+0-1+0+{1\over\sqrt2}+0-{1\over\sqrt2}+0+{1\over\sqrt3}+\cdots\\ \sum(a_n^2-b_n^2)&=1-1+1-1+{1\over2}-{1\over2}+{1\over2}-{1\over2}+{1\over3}-{1\over3}+\cdots\\ \sum2a_nb_n&=0+0+0+0+0+0+0+0+0+0+\cdots \end{align}$$
and the divergent (harmonic) sums
$$\sum a_n^2=\sum b_n^2=(0)+1+0+1+0+{1\over2}+0+{1\over2}+0+{1\over3}+0+\cdots$$
(where the "$(0)$" denotes the first term of the $b_n^2$ sum).
(Note: The remark at the end of the OP, that $a_n\ge0$, is presumably unintended, since otherwise it would combine with the convergenece of $\sum a_n$ to show that $\sum a_n^2$ also converges, which would then show that $\sum b_n^2$ converges as well.)
The answer is no. For example, take $$ a_n = \frac{(-1)^n}{\sqrt{n+1}}, \quad b_n = \frac{(-1)^{\lfloor n/2\rfloor}}{\sqrt{n+1}} $$ where $n=0,1,2,3,\dots$ Or, if you prefer, set $$ z_n = \frac{(-1)^n}{\sqrt{n+1}} + i\frac{(-1)^{\lfloor n/2\rfloor}}{\sqrt{n+1}} $$