(Given a certain application of this problem, I'm surprised I couldn't find any discussion about it specifically. I'm probably just searching for the wrong terms. In any case, this one's been bothering me for a while.)
Setup
(Unlike most "tetrominoes covering a bounded grid" problems I've seen so far,) let $M$ be the entire set of one-sided tetrominos: $$\{I,O,T,J,L,S,Z\}.$$
A "square" is a covering of a $4 \times 4$ grid with tiles in $M$, without overlapping, overhangs, or leaving any holes.
Let $n_K$ be the number of $K$-pieces in a covering, where $K \in M$. Since $4$ tetrominoes are needed to cover the grid, ($\because (4 \times 4) / 4 = 4$) it follows that for any square, $$\sum_{K \in M}n_K = n_I + n_O + n_T + n_J + n_L + n_S + n_Z = 4,$$ but this does not imply the reverse.
Here's an example of a square, where $n_I,n_L,n_J,n_Z = 1$, and $n_K = 0$ for all other $K \in M$:
$$ \begin{array}{|c|c|}
\hline
I&I&I&I \\\hline
L&J&J&J \\\hline
L&Z&Z&J \\\hline
L&L&Z&Z \\\hline
\end{array} $$
Problem
For the case of $n_T = 1$ and $\sum_{K\in M, K\neq T}n_K = 3$, does a square exist?
Put informally, if you are trying to make a $4 \times 4$ square in Tetris with whole tetriminoes, if you place a $T$-piece in it, can you successfully complete the square with other pieces or are you forced to wait for another $T$-piece?
From my absent-minded pondering, it appears the answer is no, no such square exists. Why not?
(My original observation was that a square must have an even number, including $0$, of $T$-pieces, but I then realized that the 3-alike case applies to any tetromino, so that's better suited for its own question.)
Best Answer
If you color the board like a checker board, every piece except the T covers 2 red and 2 black. The T piece covers 3 of one color. A 4x4 checker board has an equal number of red and black.