Assume that $R$ is a semisimple left square-full ring. Since ${}_RR$ is a semisimple module, ${}_RR= S_1^{(n_1)} \oplus \ldots \oplus S_k^{(n_k)}$ where $k\in \mathbb{Z}^+$, $n_1,\ldots,n_k$ are positive integers, and each $S_\ell$ is a simple left $R$-module with $S_i \not\cong S_j$ for $i\neq j$. We show, under the hypothesis that ${}_RR$ is square-full, that $n_\ell \geq 2$ for each $\ell=1,\ldots,k$. Assume on the contrary that $n_t=1$ for some $t=1,\ldots,k$. If there exists an embedding $\gamma: S_t \oplus S_t \to {}_RR$, then $S_t \oplus S_t \cong \gamma(S_t \oplus S_t) \cong \bigoplus_{\lambda\in \Lambda} S_\lambda^{(a_\lambda)}$ where $\Lambda \subset \lbrace 1,\ldots, k \rbrace$ and $a_\lambda$ are positive integers with $a_\lambda \leq n_\lambda$ for each $\lambda\in \Lambda$. Observe that $\bigoplus_{\lambda\in \Lambda} S_\lambda^{(a_\lambda)}$ has at most one copy of $S_t$ whereas $S_t \oplus S_t$ is a direct sum of two copies of $S_t$, a contradiction. Consequently, $S_t \oplus S_t$ can not embedded in ${}_RR$ which implies that $S_t$ does not contain a nonzero square-root of ${}_RR$, contradicting the hypothesis that ${}_RR$ is square-full. Thus, $n_\ell \geq 2$ for each $\ell=1,\ldots,k$.
Since $R_R$ is semisimple, then $R_R = U_1^{(b_1)} \oplus \ldots \oplus U_s^{(b_s)}$ where $s\in \mathbb{Z}^+$, $b_1,\ldots,b_s$ are positive integers, and each $U_m$ is a simple right $R$-module with $U_i \not\cong U_j$ for $i \neq j$. Put $\mathfrak{A}_\ell=\text{End}(S_\ell)$ and $\mathfrak{B}_m = \text{End}(U_m)$ for each $\ell=1,\ldots,k$ and $m=1,\ldots,s$.
Since $\lbrace S_1,\ldots,S_k \rbrace$ and $\lbrace U_1,\ldots,U_s \rbrace$ are two collections of pairwise non-isomorphic simple modules, then $\text{Hom}\bigl(S_i^{(n_i)}, S_j^{(n_j)} \bigr)=0=\text{Hom}\bigl(U_\alpha^{(b_\alpha)}, U_\beta^{(b_\beta)} \bigr)$ for $i \neq j$ and $\alpha \neq \beta$.
We now have the ring isomorphisms
\begin{align*}
R
\cong
\text{End}({}_RR)
\cong
\text{End}\bigl( S_1^{(n_1)} \oplus \ldots \oplus S_k^{(n_k)} \bigr)
\cong
\prod_{\ell=1}^k \text{End}(S^{(n_\ell)}_\ell)
\cong
\prod_{\ell=1}^k \mathbb{M}_{n_\ell}(\mathfrak{A}_\ell), \\
R
\cong
\text{End}(R_R)
\cong
\text{End}\bigl(U_1^{(b_1)} \oplus \ldots \oplus U_s^{(b_s)} \bigr)
\cong
\prod_{m=1}^s \text{End}\bigl(U^{(b_m)}_m \bigr)
\cong
\prod_{m=1}^s \mathbb{M}_{b_m}(\mathfrak{B}_m).
\end{align*}
Thus, $\prod_{\ell=1}^k \mathbb{M}_{n_\ell}(\mathfrak{A}_\ell) \cong \prod_{m=1}^s \mathbb{M}_{b_m}(\mathfrak{B}_m)$. By the uniqueness part of Wedderburn-Artin theorem, $k=s$ and $\mathfrak{B}_1,\ldots,\mathfrak{B}_s$ can be reindexed so that $b_t=n_t \geq 2$ for each $t=1,\ldots,k$. \
Now, let $X$ be a nonzero submodule of $R_R=U_1^{(b_1)} \oplus \ldots \oplus U_k^{(b_k)}$. Since $X$ is a summand of $R_R$, there exist $\Omega \subset \lbrace 1,\ldots,k \rbrace$ and positive integers $d_1,\ldots,d_k$ with $d_\ell \leq b_\ell$, for each $\ell=1,\ldots,k$, such that $X \cong \bigoplus_{\omega \in \Omega} U_\omega^{(d_\omega)}$. Let $Y$ be a simple submodule of $X$. Thus, $Y \cong U_\omega$ for some $\omega\in \Omega$ and hence $Y \oplus Y \cong U_\omega \oplus U_\omega \hookrightarrow R_R$ (as $b_\omega \geq 2$). That is to say $Y$ is a square-root in $R_R$ that is contained in $X$. Hence, as $X_R \subseteq R_R$ is arbitrary, $R_R$ is square-full. i.e. $R$ is right square-full. Analogously, $R$ is left square-full if $R$ is right square-full.
