The base in which $1065 = 13 \cdot 54$ is true

Since $3\cdot4-5$ ends in a zero, your base divides $12-5=7$. Therefore, your base is either $1$ or $7$. Base $1$ is rather silly and usually avoided (see the comments for a discussion for how to interpret base $1$).

To give some details, if $r$ is the base, then, in base $10$, $13_r=1\cdot r+3$ and $54_r=5r+4$. Therefore, by the distributive law, $$ 13_r\cdot54_r=(r+3)(5r+4)=5r^2+4r+15r+12\equiv12\pmod{r}. $$ On the other hand, in base $10$, $$ 1065_r=1\cdot r^3+0\cdot r^2+6r+5\equiv 5\pmod{r}. $$ Therefore, for equality, it must be that $5\equiv 12\pmod{r}.$.

The base is clearly less than $11$, since $13\cdot54\lt20\cdot54=T80\lt1000$ (where $T$ stands for Ten) in any base greater than $10$. Because of the "$6$" in "$1065$," the base must presumably* be at least $7$. But the base can't be even, since $1065$ is odd in any even base, while $54$ is even. That leaves only $9$ and $7$ as possible bases. But $9$ doesn't work since $1065$ is not divisible by $3$ in base $9$, while $13$ is. So we're left with base $7$, which does indeed work.

*It's purely convention, based on the desire to have a unique expression for each number, that leads us to assume that a number written in base $r$ can only be written with symbols for digits up to $r-1$. (E.g., we could add the extra digit $T$ to base ten, but then we'd have duplications such as $1T=20$.) Note, the OP's approach, solving $r^3+6r+5=(r+3)(5r+4)$ for $r$, is agnostic with respect to this convention. If you reject the convention in this answer's approach, you merely need rule out the other odd possibilities less than $11$, $r=5$ and $r=3$. For $r=5$, the left hand side would be a multiple of $5$ but not the right; for $r=3$, the left hand side is not a multiple of $3$ but $13$ is.