Consider a a diagonal $n\times n$ matrix $M$ with real diagonal elements $m_{11},m_{22},\dots,m_{nn}>0$. I am interested in finding positive-definite $n \times n$ matrices $X$ that solve \begin{equation}M=X^2\,.\tag*{(1)}\end{equation} An obvious solution $X=diag(\sqrt{m_{11}},\sqrt{m_{22}},\dots,\sqrt{m_{nn}})$. Is that the only positive-definite solution?
If I do not restrict my search to positive-definite $M$ and $X$, then I can find $2\times 2$ $M$ for which non-diagonal $X$ solve equation (1). There are plenty of examples on wikipedia. All examples listed there are not positive-definite. I haven't been able to come up with a positive-definite example, thus my hypothesis that $X=diag(\sqrt{m_{11}},\sqrt{m_{22}},\dots,\sqrt{m_{nn}})$ is the only positive-definite solution for positive-definite $M$.
Among all square roots of a positive semidefinite matrix $M$, exactly one of them is positive semidefinite. This fact is usually summarised as "every positive semidefinite matrix has a unique positive semidefinite square root". This positive semidefinite square root is usually denoted by $M^{1/2}$ or $\sqrt{M}$.
In your case, since $M=\operatorname{diag}(m_{11},\ldots,m_{nn})$ is positive semidefinite and $X=\operatorname{diag}(\sqrt{m_{11}},\ldots,\sqrt{m_{nn}})$ is already a positive semidefinite square root of $M$, the only positive semidefinite square root of $M$ is $X$. So, the answer to the question in the title is no, $M$ has not any non-diagonal positive definite square root.