$\definecolor{A}{RGB}{121,96,253}\color{A}\fbox{Solution}$ Neither $a_1=45$ nor $a_2=2700$ is a perfect square. Observe, that $(a_n)_{n\ge 1}$ is an increasing sequence and for all $m\ge 2$ we have $a_m\ge a_2>15a_1>49$. Therefore $$(a_m+7)^2=a_m^2+14a_m+49<a_m^2+15a_m=a_{m+1}$$and $$a_{m+1}=a_m^2+15a_m<a_m^2+16a_m+64=(a_m+8)^2$$Observe $$(a_m+7)^2<a_{m+1}<(a_m+8)^2\implies a_{m+1}\text{ is not a perfect square}.\blacksquare$$ #1871

Write the sequence as $a_n=a_{n-1}(a_{n-1}+15)$ and observe that the sequence is strictly increasing due to the first term being positive and then subsequent addition of two positive term. The gcd of $y^{2}=a_{n-1}(a_{n-1}+15)$ is $15$ Now, dividing the expression from $15^{2}$ we get $(\frac{y}{15})^{2}=(\frac{a_{n-1}}{15})(\frac{a_{n-1}}{15}+1)$ Since two consecutive numbers cannot be perfect squares we get a contradiction.

OGGY_666 wrote: So, for this sequence to contain perfect squares we must have that either the factors are either multiple of each other(which is clearly not possible) or both $a_{n-1}$ and $a_{n-1}+15$ are perfect squares. Not true. Consider numbers $5^3, 5\cdot 7^2$. Their product is a square, but none is divisible by the other and none is multiple the second.