Solution:

Suppose on the contrary that it is possible to cover the board with 16 strominos such that each unit square is covered by either one or two strominos. If there are $k$ squares that are covered by exactly one stromino then $2(25-k)+k=16 \times 3=48$ and hence $k=2$. Thus there are exactly two squares which are covered by only one stromino.

We colour the board with three colours red, blue, green as follows. The square corresponding to the $i$-th row and the $j$-th column is coloured red if $i+j \equiv 0 \pmod{3}$, green if $i+j \equiv 1 \pmod{3}$ and blue otherwise. Then there are 9 red squares, 8 green squares and 8 blue squares. Note that each stromino covers exactly one square of each colour. Therefore the two squares that are covered by only one stromino are both red. For each such square $i+j \equiv 0 \pmod{3}$ where $i$ and $j$ are its row and column number.

We now colour the board with a different scheme. We colour the square corresponding to the $i$-th row and the $j$-th column red if $i-j \equiv 0 \pmod{3}$, green if $i-j \equiv 1 \pmod{3}$ and blue otherwise. Again, there are 9 red squares and hence the two squares covered by only one stromino are both red. For each such square $i-j \equiv 0 \pmod{3}$ where $i$ and $j$ are its row and column number. Thus, each of the two squares covered by only one stromino satisfies $i+j \equiv 0 \pmod{3}$ and $i-j \equiv 0 \pmod{3}$ where $i$ and $j$ are its row and column number. This implies that $i=j=3$. This is a contradiction because there is only one such square.